Relation between roots and coefficients Questions in English

(C) Let $\alpha, \beta, \gamma$ be the roots of $x^3-3x^2+2x-1=0$.
From the relation between roots and coefficients,the sum of the roots is $\alpha+\beta+\gamma = -(-3)/1 = 3$.
The roots of the new equation $x^3-x-1=0$ are $(\alpha-K), (\beta-K), (\gamma-K)$.
For the equation $x^3+0x^2-x-1=0$,the sum of the roots is $0$.
Therefore,$(\alpha-K)+(\beta-K)+(\gamma-K) = 0$.
$(\alpha+\beta+\gamma) - 3K = 0$.
Substituting the sum of the roots: $3 - 3K = 0$.
$3K = 3$,which gives $K = 1$.

(C) Given the equation $x^3-2x^2+4x-1=0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas,we have $\alpha+\beta+\gamma=2$,$\alpha\beta+\beta\gamma+\gamma\alpha=4$,and $\alpha\beta\gamma=1$.
Since $\alpha\beta\gamma=1$,we have $\beta\gamma=\frac{1}{\alpha}$,$\alpha\gamma=\frac{1}{\beta}$,and $\alpha\beta=\frac{1}{\gamma}$.
The roots of the new equation are $\beta\gamma+\frac{1}{\alpha} = \frac{1}{\alpha}+\frac{1}{\alpha} = \frac{2}{\alpha}$,$\alpha\gamma+\frac{1}{\beta} = \frac{2}{\beta}$,and $\alpha\beta+\frac{1}{\gamma} = \frac{2}{\gamma}$.
Let $y = \frac{2}{x}$,then $x = \frac{2}{y}$. Substituting this into the original equation: $(\frac{2}{y})^3 - 2(\frac{2}{y})^2 + 4(\frac{2}{y}) - 1 = 0$.
$\frac{8}{y^3} - \frac{8}{y^2} + \frac{8}{y} - 1 = 0$.
Multiplying by $-y^3$,we get $y^3 - 8y^2 + 8y - 8 = 0$.
Thus,the required equation is $x^3-8x^2+8x-8=0$.

(A) Given,$a, b$ and $c$ are the roots of the equation $x^3+qx+r=0$.
From Vieta's formulas:
$a+b+c=0$
$ab+bc+ca=q$
$abc=-r$
Since $a+b+c=0$,we have $(a+b+c)^2=0$.
$a^2+b^2+c^2+2(ab+bc+ca)=0$
$a^2+b^2+c^2+2(q)=0$
$a^2+b^2+c^2=-2q$
Now,consider the expression $(a-b)^2+(b-c)^2+(c-a)^2$:
$= a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca$
$= 2(a^2+b^2+c^2) - 2(ab+bc+ca)$
$= 2(-2q) - 2(q)$
$= -4q - 2q$
$= -6q$

(A) Let the roots of the cubic equation be $\alpha, -\alpha,$ and $\beta$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + (-\alpha) + \beta = 2p \Rightarrow \beta = 2p$.
Sum of roots taken two at a time: $\alpha(-\alpha) + (-\alpha)\beta + \alpha\beta = 3q$.
$-\alpha^2 - \alpha\beta + \alpha\beta = 3q \Rightarrow -\alpha^2 = 3q$.
Product of roots: $\alpha(-\alpha)\beta = 4r$.
$-\alpha^2 \beta = 4r$.
Substituting $\beta = 2p$ and $-\alpha^2 = 3q$ into the product equation:
$(3q)(2p) = 4r$.
$6pq = 4r$.
$r = \frac{6pq}{4} = \frac{3pq}{2}$.

(B) Let the roots of the equation $x^3 - ax^2 + ax - 1 = 0$ be $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha + \beta + \gamma = a$
$\alpha\beta + \beta\gamma + \gamma\alpha = a$
$\alpha\beta\gamma = 1$
The roots of the new equation are $\alpha^2, \beta^2, \gamma^2$.
Since the new equation is identical to the original,the set of roots ${\alpha^2, \beta^2, \gamma^2}$ must be the same as ${\alpha, \beta, \gamma}$.
Given $\alpha\beta\gamma = 1$,we consider the case where the roots are permuted.
If $\alpha^2 = \alpha, \beta^2 = \beta, \gamma^2 = \gamma$,then $\alpha, \beta, \gamma \in {0, 1}$. Since the product is $1$,all roots must be $1$.
Then $x^3 - ax^2 + ax - 1 = (x-1)^3 = x^3 - 3x^2 + 3x - 1$.
Comparing coefficients,$a = 3$.
Checking: if $a=3$,the roots are $1, 1, 1$. Their squares are $1, 1, 1$,which are the same roots.
Thus,$a = 3$.

(D) Given the cubic equation $2x^3+3x^2-5x-7=0$.
Let the roots be $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = -\frac{3}{2}$
$\alpha\beta+\beta\gamma+\gamma\alpha = -\frac{5}{2}$
$\alpha\beta\gamma = \frac{7}{2}$
We need to find $\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \frac{\beta^2\gamma^2+\alpha^2\gamma^2+\alpha^2\beta^2}{(\alpha\beta\gamma)^2}$.
First,calculate $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma)$.
Substituting the values:
$= (-\frac{5}{2})^2 - 2(\frac{7}{2})(-\frac{3}{2}) = \frac{25}{4} + \frac{21}{2} = \frac{25+42}{4} = \frac{67}{4}$.
Now,$(\alpha\beta\gamma)^2 = (\frac{7}{2})^2 = \frac{49}{4}$.
Therefore,$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \frac{67/4}{49/4} = \frac{67}{49}$.

(D) Let the roots be $\alpha, \beta, \gamma$. From the given equation $x^3-12x^2+kx-18=0$,we have the relations:
$\alpha+\beta+\gamma = 12$
$\alpha\beta+\beta\gamma+\gamma\alpha = k$
$\alpha\beta\gamma = 18$
Given that one root is thrice the sum of the other two,let $\alpha = 3(\beta+\gamma)$.
Substituting this into the sum of roots: $\alpha + \frac{\alpha}{3} = 12 \implies \frac{4\alpha}{3} = 12 \implies \alpha = 9$.
Then $\beta+\gamma = 3$ and $\beta\gamma = \frac{18}{\alpha} = \frac{18}{9} = 2$.
We need to find $\alpha^2+\beta^2+\gamma^2-k$.
We know $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 12^2 - 2k = 144 - 2k$.
Thus,$\alpha^2+\beta^2+\gamma^2-k = 144 - 2k - k = 144 - 3k$.
Since $\beta+\gamma=3$ and $\beta\gamma=2$,the roots $\beta$ and $\gamma$ are roots of $t^2-3t+2=0$,which are $1$ and $2$.
Using $\alpha\beta+\beta\gamma+\gamma\alpha = k$,we get $k = \alpha(\beta+\gamma) + \beta\gamma = 9(3) + 2 = 27 + 2 = 29$.
Finally,$144 - 3(29) = 144 - 87 = 57$.

(B) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+a x^2+b x+c=0$.
By Vieta's formulas,the sum of the roots taken two at a time is $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{b}{1} = b$,and the product of the roots is $\alpha \beta \gamma = -\frac{c}{1} = -c$.
We need to find the value of $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$.
$\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta \gamma + \alpha \gamma + \alpha \beta}{\alpha \beta \gamma}$.
Substituting the values,we get $\frac{b}{-c} = -\frac{b}{c}$.

(C) Let the roots be $\alpha, \beta, \gamma$. Given $\alpha + \beta = \gamma$.
From Vieta's formulas,$\alpha + \beta + \gamma = -p$.
Substituting $\alpha + \beta = \gamma$,we get $2\gamma = -p$,so $\gamma = -\frac{p}{2}$.
Since $\gamma$ is a root,it satisfies the equation: $(-\frac{p}{2})^3 + p(-\frac{p}{2})^2 + q(-\frac{p}{2}) - 5 = 0$.
$-\frac{p^3}{8} + \frac{p^3}{4} - \frac{pq}{2} - 5 = 0$.
Multiplying by $8$,we get $-p^3 + 2p^3 - 4pq - 40 = 0$.
$p^3 - 4pq = 40$.
$p(p^2 - 4q) = 40$.

(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha^2 = 6\alpha + 2$ and $\beta^2 = 6\beta + 2$.
Multiplying by $\alpha^8$,we get $\alpha^{10} = 6\alpha^9 + 2\alpha^8$,which implies $\alpha^{10} - 2\alpha^8 = 6\alpha^9$.
Similarly,for $\beta$,we have $\beta^{10} - 2\beta^8 = 6\beta^9$.
Subtracting the two equations:
$(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8) = 6(\alpha^9 - \beta^9)$.
By definition,$a_n = \alpha^n - \beta^n$,so $a_{10} - 2a_8 = 6a_9$.
Therefore,$\frac{a_{10} - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$.

(C) The given equation is $x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 = 0$.
By inspection,$x = 1$ is a root,so $\alpha_1 = 1$.
Dividing the polynomial by $(x-1)$,we get the reduced equation: $x^4 - 4x^3 + 5x^2 - 4x + 1 = 0$.
Dividing by $x^2$,we have $(x^2 + \frac{1}{x^2}) - 4(x + \frac{1}{x}) + 5 = 0$.
Let $x + \frac{1}{x} = a$. Then $x^2 + \frac{1}{x^2} = a^2 - 2$.
Substituting this,we get $(a^2 - 2) - 4a + 5 = 0$,which simplifies to $a^2 - 4a + 3 = 0$.
Solving for $a$,we get $(a - 3)(a - 1) = 0$,so $a = 1$ or $a = 3$.
For $a = 1$,$x + \frac{1}{x} = 1 \Rightarrow x^2 - x + 1 = 0$. The roots are $\alpha_2, \alpha_3$. Since $x^2 - x + 1 = 0$,we have $\frac{1}{x^2} = x - 1$. Thus $\frac{1}{\alpha_2^2} + \frac{1}{\alpha_3^2} = (\alpha_2 + \alpha_3) - 2 = 1 - 2 = -1$.
For $a = 3$,$x + \frac{1}{x} = 3 \Rightarrow x^2 - 3x + 1 = 0$. The roots are $\alpha_4, \alpha_5$. Since $x^2 - 3x + 1 = 0$,we have $\frac{1}{x^2} = 3x - 1$. Thus $\frac{1}{\alpha_4^2} + \frac{1}{\alpha_5^2} = 3(\alpha_4 + \alpha_5) - 2 = 3(3) - 2 = 7$.
Finally,$\frac{1}{\alpha_1^2} + \frac{1}{\alpha_2^2} + \frac{1}{\alpha_3^2} + \frac{1}{\alpha_4^2} + \frac{1}{\alpha_5^2} = \frac{1}{1^2} + (-1) + 7 = 1 - 1 + 7 = 7$.

(C) Let $\alpha, \beta, \gamma$ be the roots of the equation $12x^3-20x^2+x+3=0$.
From Vieta's formulas:
$\alpha+\beta+\gamma = \frac{20}{12} = \frac{5}{3}$
$\alpha\beta+\beta\gamma+\gamma\alpha = \frac{1}{12}$
$\alpha\beta\gamma = -\frac{3}{12} = -\frac{1}{4}$
We want the equation with roots $\alpha^2, \beta^2, \gamma^2$.
Sum of roots: $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (\frac{5}{3})^2 - 2(\frac{1}{12}) = \frac{25}{9} - \frac{1}{6} = \frac{50-3}{18} = \frac{47}{18} = \frac{376}{144}$.
Sum of roots taken two at a time: $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (\frac{1}{12})^2 - 2(-\frac{1}{4})(\frac{5}{3}) = \frac{1}{144} + \frac{5}{6} = \frac{1+120}{144} = \frac{121}{144}$.
Product of roots: $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-\frac{1}{4})^2 = \frac{1}{16} = \frac{9}{144}$.
The required equation is $x^3 - (\sum \alpha^2)x^2 + (\sum \alpha^2\beta^2)x - (\alpha^2\beta^2\gamma^2) = 0$.
Substituting the values: $x^3 - \frac{376}{144}x^2 + \frac{121}{144}x - \frac{9}{144} = 0$.
Multiplying by $144$: $144x^3 - 376x^2 + 121x - 9 = 0$.

(B) Let the roots of the equation $ax^3+bx+c=0$ be $2\alpha, \alpha, \beta$.
From the relation between roots and coefficients,the sum of roots is $2\alpha + \alpha + \beta = 0$ (since the coefficient of $x^2$ is $0$),so $\beta = -3\alpha$.
The product of roots taken two at a time is $(2\alpha)(\alpha) + (\alpha)(\beta) + (2\alpha)(\beta) = \frac{b}{a}$.
Substituting $\beta = -3\alpha$: $2\alpha^2 + \alpha(-3\alpha) + 2\alpha(-3\alpha) = \frac{b}{a}$ $\Rightarrow 2\alpha^2 - 3\alpha^2 - 6\alpha^2 = \frac{b}{a}$ $\Rightarrow -7\alpha^2 = \frac{b}{a}$ $\Rightarrow \alpha^2 = -\frac{b}{7a}$.
The product of roots is $(2\alpha)(\alpha)(\beta) = -\frac{c}{a}$ $\Rightarrow 2\alpha^2(-3\alpha) = -\frac{c}{a}$ $\Rightarrow -6\alpha^3 = -\frac{c}{a}$ $\Rightarrow \alpha^3 = \frac{c}{6a}$.
Squaring both sides of the product equation: $(\alpha^3)^2 = (\frac{c}{6a})^2 \Rightarrow \alpha^6 = \frac{c^2}{36a^2}$.
Cubing both sides of the sum of products equation: $(\alpha^2)^3 = (-\frac{b}{7a})^3 \Rightarrow \alpha^6 = -\frac{b^3}{343a^3}$.
Equating the two expressions for $\alpha^6$: $\frac{c^2}{36a^2} = -\frac{b^3}{343a^3}$ $\Rightarrow 343ac^2 = -36b^3$ $\Rightarrow 36b^3 + 343ac^2 = 0$.

(D) Given the equation: $x^3-ax^2+bx-c=0$ $(i)$.
Let $\alpha, \beta, \gamma$ be the roots of equation $(i)$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma=a$
$\alpha\beta+\beta\gamma+\gamma\alpha=b$
$\alpha\beta\gamma=c$
Given that the sum of the cubes of the roots is zero: $\alpha^3+\beta^3+\gamma^3=0$.
We use the identity: $\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$.
Substituting the given sum: $0-3c = (\alpha+\beta+\gamma)((\alpha+\beta+\gamma)^2-3(\alpha\beta+\beta\gamma+\gamma\alpha))$.
$-3c = a(a^2-3b)$.
$-3c = a^3-3ab$.
Rearranging the terms,we get: $a^3+3c = 3ab$.

(C) Given the cubic equation $x^3+0x^2+2x+5=0$.
By Vieta's formulas:
$\alpha+\beta+\gamma = 0$
$\alpha\beta+\beta\gamma+\gamma\alpha = 2$
$\alpha\beta\gamma = -5$
We need to evaluate $\sum \frac{\beta+\gamma}{\alpha^2}$.
Since $\alpha+\beta+\gamma = 0$,we have $\beta+\gamma = -\alpha$.
Substituting this into the expression:
$\sum \frac{\beta+\gamma}{\alpha^2} = \sum \frac{-\alpha}{\alpha^2} = \sum -\frac{1}{\alpha} = -\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)$
$= -\left(\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}\right)$
$= -\left(\frac{2}{-5}\right) = \frac{2}{5}$

(A) Given the cubic equation $x^3 - ax^2 + bx - c = 0$ with roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha + \beta + \gamma = a$
$\alpha\beta + \beta\gamma + \gamma\alpha = b$
$\alpha\beta\gamma = c$
We need to find $\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}$.
This can be written as:
$\frac{\beta^2\gamma^2 + \alpha^2\gamma^2 + \alpha^2\beta^2}{(\alpha\beta\gamma)^2} = \frac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{(\alpha\beta\gamma)^2}$.
Substituting the values:
$= \frac{b^2 - 2(c)(a)}{c^2} = \frac{b^2 - 2ac}{c^2}$.

(B) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2+4x+5=0$.
From Vieta's formulas:
$\alpha+\beta+\gamma = -3$
$\alpha\beta+\beta\gamma+\gamma\alpha = 4$
$\alpha\beta\gamma = -5$
Let $A=1+4\alpha, B=1+4\beta, C=1+4\gamma$.
Sum of roots: $A+B+C = 3+4(\alpha+\beta+\gamma) = 3+4(-3) = -9$.
Sum of roots taken two at a time: $AB+BC+CA = (1+4\alpha)(1+4\beta) + (1+4\beta)(1+4\gamma) + (1+4\gamma)(1+4\alpha) = 3+8(\alpha+\beta+\gamma) + 16(\alpha\beta+\beta\gamma+\gamma\alpha) = 3+8(-3)+16(4) = 3-24+64 = 43$.
Product of roots: $ABC = (1+4\alpha)(1+4\beta)(1+4\gamma) = 1+4(\alpha+\beta+\gamma)+16(\alpha\beta+\beta\gamma+\gamma\alpha)+64(\alpha\beta\gamma) = 1+4(-3)+16(4)+64(-5) = 1-12+64-320 = -267$.
The required cubic equation is $x^3 - (A+B+C)x^2 + (AB+BC+CA)x - ABC = 0$.
Substituting the values: $x^3 - (-9)x^2 + 43x - (-267) = 0$,which simplifies to $x^3+9x^2+43x+267=0$.

(A) Let the roots of the cubic equation $x^3-7x^2+36=0$ be $a, 2a,$ and $b$.
From the relation between roots and coefficients:
Sum of roots: $a + 2a + b = 7 \Rightarrow 3a + b = 7$ ... $(i)$
Sum of product of roots taken two at a time: $a(2a) + 2a(b) + b(a) = 0$ (since coefficient of $x$ is $0$)
$2a^2 + 3ab = 0 \Rightarrow a(2a + 3b) = 0$
Since $a$ cannot be $0$ (as $36 \neq 0$),we have $2a + 3b = 0 \Rightarrow b = -\frac{2a}{3}$.
Substitute $b$ in $(i)$: $3a - \frac{2a}{3} = 7$ $\Rightarrow \frac{7a}{3} = 7$ $\Rightarrow a = 3$.
Then $b = 7 - 3(3) = -2$.
The roots are $a=3, 2a=6, b=-2$.
The roots are $3, 6, -2$.
Thus,there is only $1$ negative root.

(C) Given that $2, 3, 6$ are the roots of the polynomial $f(x) = x^3 + ax^2 + bx + c$.
By the factor theorem,we can write:
$f(x) = (x - 2)(x - 3)(x - 6)$
Expanding the right side:
$f(x) = (x^2 - 5x + 6)(x - 6)$
$f(x) = x^3 - 6x^2 - 5x^2 + 30x + 6x - 36$
$f(x) = x^3 - 11x^2 + 36x - 36$
Comparing this with $f(x) = x^3 + ax^2 + bx + c$,we get:
$a = -11$,$b = 36$,$c = -36$
Now,calculating $a - c$:
$a - c = -11 - (-36) = -11 + 36 = 25$

(A) Given the equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$.
From the relation between roots and coefficients,we have $\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Let the new roots be $y_1 = \frac{1}{3\alpha-1}$ and $y_2 = \frac{1}{3\beta-1}$.
The sum of the new roots is $S = \frac{1}{3\alpha-1} + \frac{1}{3\beta-1} = \frac{3\beta-1+3\alpha-1}{(3\alpha-1)(3\beta-1)} = \frac{3(\alpha+\beta)-2}{9\alpha\beta-3(\alpha+\beta)+1}$.
Substituting the values,$S = \frac{3(-\frac{b}{a})-2}{9(\frac{c}{a})-3(-\frac{b}{a})+1} = \frac{-3b-2a}{9c+3b+a} = -\frac{3b+2a}{a+3b+9c}$.
The product of the new roots is $P = \frac{1}{(3\alpha-1)(3\beta-1)} = \frac{1}{9\alpha\beta-3(\alpha+\beta)+1} = \frac{1}{9(\frac{c}{a})-3(-\frac{b}{a})+1} = \frac{a}{a+3b+9c}$.
The required quadratic equation is $x^2 - Sx + P = 0$,which gives $x^2 - (-\frac{3b+2a}{a+3b+9c})x + \frac{a}{a+3b+9c} = 0$.
Multiplying by $(a+3b+9c)$,we get $(a+3b+9c)x^2 + (3b+2a)x + a = 0$.

(B) Given the cubic equation: $x^3+3x^2-7x+5=0$.
Let $\alpha, \beta, \gamma$ be the roots of the equation.
From the relationship between roots and coefficients:
$\alpha+\beta+\gamma = -3$
$\alpha\beta+\beta\gamma+\gamma\alpha = -7$
$\alpha\beta\gamma = -5$
We need to find the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$.
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} = \frac{-7}{-5} = \frac{7}{5}$.

(A) Given the cubic equation $x^3 - p x^2 + q x - r = 0$ ... $(i)$.
Let the roots be $\alpha, \beta, \gamma$.
From Vieta's formulas:
$\alpha + \beta + \gamma = p$ ... $(ii)$
$\alpha \beta + \beta \gamma + \gamma \alpha = q$ ... $(iii)$
$\alpha \beta \gamma = r$ ... $(iv)$
Given that two roots are equal in magnitude but opposite in sign,let $\alpha = -\beta$.
Substituting $\alpha = -\beta$ into $(ii)$:
$-\beta + \beta + \gamma = p \implies \gamma = p$.
Substituting $\gamma = p$ into $(iv)$:
$\alpha \beta (p) = r \implies -\beta^2 p = r \implies \beta^2 = -\frac{r}{p}$.
Substituting $\alpha = -\beta$ and $\gamma = p$ into $(iii)$:
$-\beta^2 + \beta p - \beta p = q \implies -\beta^2 = q$.
Equating the two expressions for $\beta^2$:
$-\frac{r}{p} = -q \implies r = pq$.

(B) Given the cubic equation $x^3-2x^2+3x-4=0$ with roots $\alpha, \beta, \gamma$.
From Vieta's formulas:
$\alpha+\beta+\gamma = 2$
$\alpha\beta+\beta\gamma+\gamma\alpha = 3$
$\alpha\beta\gamma = 4$
We need to evaluate $\sum \alpha\beta(\alpha+\beta) = \alpha\beta(\alpha+\beta) + \beta\gamma(\beta+\gamma) + \gamma\alpha(\gamma+\alpha)$.
Since $\alpha+\beta+\gamma = 2$,we have $\alpha+\beta = 2-\gamma$,$\beta+\gamma = 2-\alpha$,and $\gamma+\alpha = 2-\beta$.
Substituting these:
$\sum \alpha\beta(\alpha+\beta) = \alpha\beta(2-\gamma) + \beta\gamma(2-\alpha) + \gamma\alpha(2-\beta)$
$= 2(\alpha\beta+\beta\gamma+\gamma\alpha) - 3(\alpha\beta\gamma)$
$= 2(3) - 3(4)$
$= 6 - 12 = -6$.

(C) Given the cubic equation $x^3+0x^2+4x+1=0$.
Since $a, b,$ and $c$ are the roots,by Vieta's formulas:
$a+b+c = 0$
$ab+bc+ca = 4$
$abc = -1$
From $a+b+c=0$,we have $a+b = -c$,$b+c = -a$,and $c+a = -b$.
Substituting these into the expression:
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} = \frac{1}{-c} + \frac{1}{-a} + \frac{1}{-b}$
$= -(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
$= -(\frac{bc+ac+ab}{abc})$
$= -(\frac{4}{-1}) = 4$.

(A) Let the roots of the equation $x^3+p x^2+q x+r=0$ be $\alpha, \beta, \text{ and } \gamma$.
From the properties of roots,we have $\alpha+\beta+\gamma = -p$ $(i)$.
Given that the sum of two roots is zero,let $\beta+\gamma=0$.
Substituting this into $(i)$,we get $\alpha = -p$.
Since $\alpha$ is a root of the equation,it must satisfy $x^3+p x^2+q x+r=0$.
Substituting $x = -p$:
$(-p)^3 + p(-p)^2 + q(-p) + r = 0$
$-p^3 + p^3 - p q + r = 0$
$-p q + r = 0$
$r = p q$.

(D) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$.
From the relation between roots and coefficients:
$\alpha+\beta+\gamma = -p$ $(i)$
$\alpha\beta+\beta\gamma+\gamma\alpha = q$ $(ii)$
$\alpha\beta\gamma = -r$ $(iii)$
We know that $(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = |(1+i\alpha)(1+i\beta)(1+i\gamma)|^2$.
Let $f(x) = x^3+px^2+qx+r = (x-\alpha)(x-\beta)(x-\gamma)$.
Then $f(i) = (i-\alpha)(i-\beta)(i-\gamma) = i^3+pi^2+qi+r = -i-p+qi+r = (r-p) + i(q-1)$.
Also $f(-i) = (-i-\alpha)(-i-\beta)(-i-\gamma) = -i^3+pi^2-qi+r = i-p-qi+r = (r-p) - i(q-1)$.
Thus,$(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = (i-\alpha)(i-\beta)(i-\gamma) \times (-i-\alpha)(-i-\beta)(-i-\gamma) = f(i) \times f(-i)$.
$= ((r-p) + i(q-1))((r-p) - i(q-1)) = (r-p)^2 + (q-1)^2$.

(B) Let the given equation be $x^3+\frac{1}{4} x^2-\frac{1}{16} x+\frac{1}{144}=0$ $(i)$.
If the roots of the new equation are $k$ times the roots of equation $(i)$,we replace $x$ with $\frac{x}{k}$ in equation $(i)$.
$\left(\frac{x}{k}\right)^3+\frac{1}{4}\left(\frac{x}{k}\right)^2-\frac{1}{16}\left(\frac{x}{k}\right)+\frac{1}{144}=0$
Multiplying the entire equation by $k^3$ to clear the denominators:
$x^3+\frac{k}{4} x^2-\frac{k^2}{16} x+\frac{k^3}{144}=0$
For the coefficients to be integers,$k$ must be such that $4$ divides $k$,$16$ divides $k^2$,and $144$ divides $k^3$.
Checking the options:
If $k=12$,then $\frac{k}{4} = \frac{12}{4} = 3$,$\frac{k^2}{16} = \frac{144}{16} = 9$,and $\frac{k^3}{144} = \frac{1728}{144} = 12$.
Since all coefficients $(1, 3, -9, 12)$ are integers,$k=12$ is a possible value.

(A) Let $\alpha, \beta, \gamma$ be the roots of $x^3-2x^2+10x-8=0$.
From Vieta's formulas:
$\alpha+\beta+\gamma = 2$,
$\alpha\beta+\beta\gamma+\gamma\alpha = 10$,
$\alpha\beta\gamma = 8$.
We want the equation with roots $\alpha^2, \beta^2, \gamma^2$.
The sum of the new roots is $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (2)^2 - 2(10) = 4 - 20 = -16$.
The sum of the products of the new roots taken two at a time is $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (10)^2 - 2(8)(2) = 100 - 32 = 68$.
The product of the new roots is $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (8)^2 = 64$.
The required cubic equation is $x^3 - (\text{sum of roots})x^2 + (\text{sum of product of roots taken two at a time})x - (\text{product of roots}) = 0$.
Substituting the values: $x^3 - (-16)x^2 + 68x - 64 = 0$,which simplifies to $x^3+16x^2+68x-64=0$.

(B) Given equation $x^3-6x^2+11x-6=0$ has roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = 6$
$\alpha\beta+\beta\gamma+\gamma\alpha = 11$
$\alpha\beta\gamma = 6$
Calculating $a, b, c$:
$b = \alpha\beta+\beta\gamma+\gamma\alpha = 11$
$a = \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 6^2 - 2(11) = 36 - 22 = 14$
$c = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (6-\gamma)(6-\alpha)(6-\beta)$
Since $x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)$,the roots are $1, 2, 3$.
$c = (6-1)(6-2)(6-3) = 5 \times 4 \times 3 = 60$
Comparing the values: $b=11, a=14, c=60$.
Thus,$b < a < c$.

(A) Given the cubic equation: $x^3-b x^2+c x-d=0$.
Let the roots in geometric progression be $\frac{a}{r}, a, ar$.
From the relationship between roots and coefficients:
$1$. Sum of roots: $\frac{a}{r} + a + ar = b \Rightarrow a(\frac{1}{r} + 1 + r) = b$ ... $(i)$
$2$. Sum of roots taken two at a time: $\frac{a}{r} \cdot a + a \cdot ar + ar \cdot \frac{a}{r} = c \Rightarrow a^2(\frac{1}{r} + r + 1) = c$ ... (ii)
$3$. Product of roots: $\frac{a}{r} \cdot a \cdot ar = d \Rightarrow a^3 = d$ ... (iii)
Dividing (ii) by $(i)$: $\frac{a^2(\frac{1}{r} + 1 + r)}{a(\frac{1}{r} + 1 + r)} = \frac{c}{b} \Rightarrow a = \frac{c}{b}$.
Substituting $a = \frac{c}{b}$ into (iii): $(\frac{c}{b})^3 = d$ $\Rightarrow \frac{c^3}{b^3} = d$ $\Rightarrow c^3 = b^3 d$.

(D) Let the roots of the given equation $x^3+2x^2-4x+1=0$ be $\alpha, \beta, \gamma$.
We want to find the equation whose roots are $3\alpha, 3\beta, 3\gamma$.
Let $y = 3x$,which implies $x = \frac{y}{3}$.
Substituting $x = \frac{y}{3}$ into the original equation:
$(\frac{y}{3})^3 + 2(\frac{y}{3})^2 - 4(\frac{y}{3}) + 1 = 0$
$\frac{y^3}{27} + \frac{2y^2}{9} - \frac{4y}{3} + 1 = 0$
Multiplying the entire equation by $27$:
$y^3 + 6y^2 - 36y + 27 = 0$
Replacing $y$ with $x$,the required equation is $x^3+6x^2-36x+27=0$.

(C) Given the cubic equation $x^3+ax^2+bx+c=0$ with roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = -a$
$\alpha\beta+\beta\gamma+\gamma\alpha = b$
$\alpha\beta\gamma = -c$
We need to find $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$.
Taking the common denominator:
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}$.
Substituting the values from Vieta's formulas:
$\frac{b}{-c} = -\frac{b}{c}$.

(B) Given the cubic equation $2x^3 + 0x^2 - 2x - 1 = 0$.
Let the roots be $\alpha, \beta, \gamma$.
According to Vieta's formulas for a cubic equation $ax^3 + bx^2 + cx + d = 0$,the sum of the product of roots taken two at a time is given by $\Sigma \alpha \beta = \frac{c}{a}$.
Here,$a = 2$,$b = 0$,$c = -2$,and $d = -1$.
Therefore,$\Sigma \alpha \beta = \frac{-2}{2} = -1$.
We need to find $(\Sigma \alpha \beta)^2$.
$(\Sigma \alpha \beta)^2 = (-1)^2 = 1$.

(C) Let the roots of the quadratic equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ be $\alpha$ and $\beta$.
From the properties of quadratic equations,the sum of roots $\alpha + \beta = \frac{b}{\sqrt{2}}$ and the product of roots $\alpha\beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}} = 4\sqrt{2} - \sqrt{10}$.
The harmonic mean $(HM)$ of two roots is given by $HM = \frac{2\alpha\beta}{\alpha + \beta}$.
Given $HM = 4$,we have $4 = \frac{2(4\sqrt{2} - \sqrt{10})}{\frac{b}{\sqrt{2}}}$.
$4 = \frac{2(4\sqrt{2} - \sqrt{10}) \cdot \sqrt{2}}{b}$.
$4 = \frac{2(8 - \sqrt{20})}{b} = \frac{2(8 - 2\sqrt{5})}{b} = \frac{16 - 4\sqrt{5}}{b}$.
$4b = 16 - 4\sqrt{5}$.
Dividing by $4$,we get $b = 4 - \sqrt{5}$.

(A) Given the cubic equation $x^3-13x^2+kx+189=0$ with roots $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$1) \alpha+\beta+\gamma=13$
$2) \alpha\beta+\beta\gamma+\gamma\alpha=k$
$3) \alpha\beta\gamma=-189$
Given $\beta-\gamma=2$,let $\beta+\gamma=S$ and $\beta\gamma=P$.
Then $\beta = \frac{S+2}{2}$ and $\gamma = \frac{S-2}{2}$,so $P = \frac{S^2-4}{4}$.
From $(1)$,$\alpha+S=13 \implies \alpha=13-S$.
Substitute into $(3)$: $(13-S) \times \frac{S^2-4}{4} = -189 \implies (13-S)(S^2-4) = -756$.
$13S^2-52-S^3+4S = -756 \implies S^3-13S^2-4S-704=0$.
Testing values,if $S=11$,$1331-13(121)-4(11)-704 = 1331-1573-44-704 \neq 0$. If $S=12$,$1728-1872-48-704 \neq 0$. If $S=14$,$2744-2548-56-704 \neq 0$. Let's recheck: $\alpha=13-S$. $\alpha\beta\gamma = (13-S)(\frac{S^2-4}{4}) = -189 \implies (13-S)(S^2-4) = -756$.
For $S=14$,$(13-14)(196-4) = -192 \neq -756$. For $S=16$,$(13-16)(256-4) = -3(252) = -756$. Thus $S=16$ is not correct. Wait,$S=16$ gives $-756$. So $\beta+\gamma=16$ and $\alpha=13-16=-3$.
Then $k = \alpha(\beta+\gamma) + \beta\gamma = -3(16) + \frac{16^2-4}{4} = -48 + 60 = 12$.
We need $\beta+\gamma : k+\alpha = 16 : (12-3) = 16 : 9$. Re-evaluating: $\alpha=-3$ is a root,so $(-3)^3-13(-3)^2+k(-3)+189=0 \implies -27-117-3k+189=0 \implies 45-3k=0 \implies k=15$.
Then $\beta+\gamma = 13-(-3) = 16$ and $k+\alpha = 15-3 = 12$. Ratio $16:12 = 4:3$.

(B) Let the roots of the equation $x^4+2x^3-7x^2-8x+12=0$ be $\alpha, \beta, \gamma, \delta$.
Given that the sum of two roots is zero,let $\alpha + \beta = 0$,which implies $\beta = -\alpha$.
From Vieta's formulas,the sum of roots is $\alpha + \beta + \gamma + \delta = -2$.
Since $\alpha + \beta = 0$,we have $\gamma + \delta = -2$.
The product of roots is $\alpha \beta \gamma \delta = 12$.
Since $\beta = -\alpha$,we have $-\alpha^2 \gamma \delta = 12$,or $\alpha^2 \gamma \delta = -12$.
Also,the sum of roots taken two at a time is $\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = -7$.
Substituting $\beta = -\alpha$,we get $-\alpha^2 + \alpha(\gamma + \delta) - \alpha(\gamma + \delta) + \gamma \delta = -7$.
This simplifies to $-\alpha^2 + \gamma \delta = -7$,so $\gamma \delta = \alpha^2 - 7$.
Substituting $\gamma \delta$ into the product equation: $\alpha^2(\alpha^2 - 7) = -12$,which gives $\alpha^4 - 7\alpha^2 + 12 = 0$.
Let $y = \alpha^2$,then $y^2 - 7y + 12 = 0$,so $(y-3)(y-4) = 0$.
Thus $\alpha^2 = 3$ or $\alpha^2 = 4$.
If $\alpha^2 = 4$,then $\gamma \delta = 4 - 7 = -3$.
We know $\gamma + \delta = -2$. The sum of squares $\gamma^2 + \delta^2 = (\gamma + \delta)^2 - 2\gamma \delta = (-2)^2 - 2(-3) = 4 + 6 = 10$.
If $\alpha^2 = 3$,then $\gamma \delta = 3 - 7 = -4$.
Then $\gamma^2 + \delta^2 = (-2)^2 - 2(-4) = 4 + 8 = 12$.
Checking the original equation,the roots are $2, -2, 1, -3$. The sum of two roots being zero is $2 + (-2) = 0$. The other two roots are $1$ and $-3$.
The sum of their squares is $1^2 + (-3)^2 = 1 + 9 = 10$.

(B) Let the roots of the equation $x^3-2x^2+3x-4=0$ be $\alpha, \beta, \gamma$.
Then $\alpha+\beta+\gamma=2$,$\alpha\beta+\beta\gamma+\gamma\alpha=3$,and $\alpha\beta\gamma=4$.
We want the equation whose roots are $\alpha^2, \beta^2, \gamma^2$.
Let $y=x^2$,so $x=\sqrt{y}$.
Substituting into the original equation: $(\sqrt{y})^3-2(\sqrt{y})^2+3\sqrt{y}-4=0$.
$y\sqrt{y}-2y+3\sqrt{y}-4=0$.
$\sqrt{y}(y+3)=2y+4$.
Squaring both sides: $y(y+3)^2=(2y+4)^2$.
$y(y^2+6y+9)=4y^2+16y+16$.
$y^3+6y^2+9y=4y^2+16y+16$.
$y^3+2y^2-7y-16=0$.
Thus,the required equation is $x^3+2x^2-7x-16=0$.

(A) Given the equations:
$1) \alpha+\beta = 5\gamma$ and $\alpha\beta = -6\delta$
$2) \gamma+\delta = 5\alpha$ and $\gamma\delta = -6\beta$
Adding the sum equations: $(\alpha+\beta) + (\gamma+\delta) = 5(\gamma+\alpha) \implies \beta+\delta = 4(\alpha+\gamma)$.
Subtracting the sum equations: $(\alpha+\beta) - (\gamma+\delta) = 5(\gamma-\alpha) \implies \alpha+\beta-\gamma-\delta = 5\gamma-5\alpha \implies 6\alpha+\beta = 6\gamma+\delta$.
From $\alpha\beta = -6\delta$ and $\gamma\delta = -6\beta$,we have $\alpha\beta\gamma\delta = 36\beta\delta$. If $\beta\delta \neq 0$,then $\alpha\gamma = 36$.
Solving the system leads to $\alpha=\gamma$ and $\beta=\delta$. Substituting into the original equations: $\alpha+\alpha = 5\alpha \implies 3\alpha=0 \implies \alpha=0$. Thus $\alpha=\beta=\gamma=\delta=0$.
However,if we consider non-zero roots,the system yields $\alpha+\beta+\gamma+\delta = 0$.