Mix Examples - Polynomials Questions in English

(A) Let $f(x) = x^{3}-6x^{2}+11x-6$ be the given polynomial.
By the Factor Theorem,we test the factors of the constant term $-6$,which are $\pm 1, \pm 2, \pm 3, \pm 6$.
For $x = 1$,$f(1) = (1)^{3}-6(1)^{2}+11(1)-6 = 1-6+11-6 = 0$. Thus,$(x-1)$ is a factor.
For $x = 2$,$f(2) = (2)^{3}-6(2)^{2}+11(2)-6 = 8-24+22-6 = 0$. Thus,$(x-2)$ is a factor.
For $x = 3$,$f(3) = (3)^{3}-6(3)^{2}+11(3)-6 = 27-54+33-6 = 0$. Thus,$(x-3)$ is a factor.
Since the polynomial is of degree $3$,it can have at most $3$ linear factors.
Therefore,the factors of $x^{3}-6x^{2}+11x-6$ are $(x-1)(x-2)(x-3)$.

(B) Let $f(x) = x^{3} + x^{2} - 4x - 4$.
We can factorise by grouping terms:
$f(x) = (x^{3} + x^{2}) - (4x + 4)$
Factor out the common terms from each group:
$f(x) = x^{2}(x + 1) - 4(x + 1)$
Now,take $(x + 1)$ as a common factor:
$f(x) = (x + 1)(x^{2} - 4)$
Using the algebraic identity $a^{2} - b^{2} = (a - b)(a + b)$,we can factor $x^{2} - 4$ as $(x - 2)(x + 2)$:
$f(x) = (x + 1)(x - 2)(x + 2)$
Thus,the factors are $(x - 2)(x + 1)(x + 2)$.

(D) Let $f(x) = 3x^{3}-x^{2}-3x+1$ be the given polynomial.
We can factorize this by grouping terms:
$f(x) = (3x^{3}-3x) - (x^{2}-1)$
$f(x) = 3x(x^{2}-1) - 1(x^{2}-1)$
$f(x) = (3x-1)(x^{2}-1)$
Since $x^{2}-1$ is a difference of squares,we can write it as $(x-1)(x+1)$.
Therefore,$f(x) = (3x-1)(x-1)(x+1)$.

(D) To evaluate $103^{3}$,we can express it as $(100 + 3)^{3}$.
Using the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$,where $a = 100$ and $b = 3$:
$(100 + 3)^{3} = (100)^{3} + (3)^{3} + 3(100)(3)(100 + 3)$
$= 1000000 + 27 + 900(103)$
$= 1000000 + 27 + 92700$
$= 1092727$

(A) We can write the expression as:
$101 \times 102 = (100 + 1)(100 + 2)$
Using the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $x = 100$,$a = 1$,and $b = 2$:
$(100 + 1)(100 + 2) = (100)^2 + (1 + 2)(100) + (1)(2)$
$= 10000 + (3)(100) + 2$
$= 10000 + 300 + 2$
$= 10302$

(B) We can express $999$ as $(1000 - 1)$.
Using the algebraic identity $(a - b)^{2} = a^{2} - 2ab + b^{2}$,where $a = 1000$ and $b = 1$:
$(999)^{2} = (1000 - 1)^{2}$
$= (1000)^{2} - 2 \times (1000) \times (1) + (1)^{2}$
$= 1000000 - 2000 + 1$
$= 998000 + 1$
$= 998001$

(A) We have the expression $4 x^{2}+20 x+25$.
This can be written in the form $a^{2}+2 a b+b^{2}$ as follows:
$4 x^{2}+20 x+25 = (2 x)^{2}+2(2 x)(5)+(5)^{2}$
Using the algebraic identity $a^{2}+2 a b+b^{2}=(a+b)^{2}$,where $a = 2x$ and $b = 5$,we get:
$(2 x+5)^{2}$
Thus,the factors are $(2 x+5)(2 x+5)$.

(N/A) We have the expression $9 y^{2}-66 y z+121 z^{2}$.
This can be written in the form $a^{2}-2 a b+b^{2} = (a-b)^{2}$.
$9 y^{2}-66 y z+121 z^{2} = (3 y)^{2} - 2(3 y)(11 z) + (11 z)^{2}$.
Comparing this with the identity $(a-b)^{2} = a^{2}-2 a b+b^{2}$,where $a = 3 y$ and $b = 11 z$.
Therefore,$9 y^{2}-66 y z+121 z^{2} = (3 y - 11 z)^{2}$.
This can be written as $(3 y - 11 z)(3 y - 11 z)$.

We use the algebraic identity $a^{2}-b^{2}=(a+b)(a-b)$,where $a = \left(2x + \frac{1}{3}\right)$ and $b = \left(x - \frac{1}{2}\right)$.
Substituting these into the identity:
$= \left[\left(2x + \frac{1}{3}\right) + \left(x - \frac{1}{2}\right)\right] \left[\left(2x + \frac{1}{3}\right) - \left(x - \frac{1}{2}\right)\right]$
Simplify the terms inside the brackets:
$= \left(2x + x + \frac{1}{3} - \frac{1}{2}\right) \left(2x - x + \frac{1}{3} + \frac{1}{2}\right)$
$= \left(3x + \frac{2-3}{6}\right) \left(x + \frac{2+3}{6}\right)$
$= \left(3x - \frac{1}{6}\right) \left(x + \frac{5}{6}\right)$

(N/A) To factorise the quadratic expression $9 x^{2}-12 x+3$,we use the splitting the middle term method.
First,we find two numbers whose product is $9 \times 3 = 27$ and whose sum is $-12$.
These two numbers are $-9$ and $-3$.
Now,rewrite the middle term $-12 x$ as $-9 x - 3 x$:
$9 x^{2}-9 x-3 x+3$
Group the terms:
$=9 x(x-1)-3(x-1)$
Factor out the common binomial $(x-1)$:
$=(9 x-3)(x-1)$
Finally,factor out the common constant $3$ from the first binomial:
$=3(3 x-1)(x-1)$

(A) We have the expression $9x^{2}-12x+4$.
This can be written in the form $a^{2}-2ab+b^{2}$ where $a=3x$ and $b=2$.
$9x^{2}-12x+4 = (3x)^{2} - 2(3x)(2) + (2)^{2}$.
Using the algebraic identity $a^{2}-2ab+b^{2} = (a-b)^{2}$,we get:
$(3x-2)^{2} = (3x-2)(3x-2)$.

(A) To expand the expression $(4a - b + 2c)^2$,we use the algebraic identity:
$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Here,$x = 4a$,$y = -b$,and $z = 2c$.
Substituting these values into the identity:
$(4a - b + 2c)^2 = (4a)^2 + (-b)^2 + (2c)^2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a)$
Calculating each term:
$= 16a^2 + b^2 + 4c^2 - 8ab - 4bc + 16ac$

(A) To expand $(3a - 5b - c)^2$,we use the algebraic identity:
$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Here,$x = 3a$,$y = -5b$,and $z = -c$.
Substituting these values into the identity:
$(3a - 5b - c)^2 = (3a)^2 + (-5b)^2 + (-c)^2 + 2(3a)(-5b) + 2(-5b)(-c) + 2(-c)(3a)$
Calculating each term:
$= 9a^2 + 25b^2 + c^2 - 30ab + 10bc - 6ca$

(A) We use the algebraic identity: $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Here,$a = -x$,$b = 2y$,and $c = -3z$.
Substituting these values into the identity:
$(-x+2y-3z)^{2} = (-x)^{2} + (2y)^{2} + (-3z)^{2} + 2(-x)(2y) + 2(2y)(-3z) + 2(-3z)(-x)$
$= x^{2} + 4y^{2} + 9z^{2} - 4xy - 12yz + 6xz$.

(A) We have the expression: $9 x^{2}+4 y^{2}+16 z^{2}+12 x y-16 y z-24 x z$.
This can be written in the form $a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a = (a+b+c)^{2}$.
Here,$a = 3x$,$b = 2y$,and $c = -4z$.
Substituting these values:
$9 x^{2}+4 y^{2}+16 z^{2}+12 x y-16 y z-24 x z = (3 x)^{2}+(2 y)^{2}+(-4 z)^{2}+2(3 x)(2 y)+2(2 y)(-4 z)+2(-4 z)(3 x)$.
Using the identity $(a+b+c)^{2}$,we get:
$(3 x+2 y-4 z)^{2}$.
Thus,the factorised form is $(3 x+2 y-4 z)(3 x+2 y-4 z)$.

(N/A) The given expression is $25 x^{2}+16 y^{2}+4 z^{2}-40 x y+16 y z-20 x z$.
We use the algebraic identity: $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Comparing the given expression with the identity,we can rewrite it as:
$=(-5 x)^{2}+(4 y)^{2}+(2 z)^{2}+2(-5 x)(4 y)+2(4 y)(2 z)+2(2 z)(-5 x)$.
Here,$a = -5x$,$b = 4y$,and $c = 2z$.
Therefore,the expression simplifies to $(-5 x+4 y+2 z)^{2}$ or $(5 x-4 y-2 z)^{2}$.

(N/A) The given expression is $16 x^{2}+4 y^{2}+9 z^{2}-16 x y-12 y z+24 x z$.
We can rewrite this expression using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$.
Here,we observe the terms:
$16 x^{2} = (4 x)^{2}$
$4 y^{2} = (-2 y)^{2}$
$9 z^{2} = (3 z)^{2}$
Now,checking the cross terms:
$2(4 x)(-2 y) = -16 x y$
$2(-2 y)(3 z) = -12 y z$
$2(3 z)(4 x) = 24 x z$
Thus,the expression becomes:
$(4 x)^{2}+(-2 y)^{2}+(3 z)^{2}+2(4 x)(-2 y)+2(-2 y)(3 z)+2(3 z)(4 x)$
Applying the identity,we get:
$(4 x-2 y+3 z)^{2}$
Therefore,the factorised form is $(4 x-2 y+3 z)(4 x-2 y+3 z)$.

(B) We know the algebraic identity:
$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$
Given that $a+b+c = 9$ and $ab+bc+ca = 26$.
Substituting these values into the identity:
$(9)^{2} = (a^{2}+b^{2}+c^{2}) + 2(26)$
$81 = (a^{2}+b^{2}+c^{2}) + 52$
Subtracting $52$ from both sides:
$a^{2}+b^{2}+c^{2} = 81 - 52$
$a^{2}+b^{2}+c^{2} = 29$

(A) To expand the expression $(3a - 2b)^3$,we use the algebraic identity:
$(x - y)^3 = x^3 - y^3 - 3xy(x - y)$
Here,$x = 3a$ and $y = 2b$.
Substituting these values into the identity:
$(3a - 2b)^3 = (3a)^3 - (2b)^3 - 3(3a)(2b)(3a - 2b)$
Calculate the cubes:
$(3a)^3 = 27a^3$
$(2b)^3 = 8b^3$
Now,simplify the expression:
$= 27a^3 - 8b^3 - 18ab(3a - 2b)$
Distribute $-18ab$ into the parentheses:
$= 27a^3 - 8b^3 - (18ab \times 3a) + (18ab \times 2b)$
$= 27a^3 - 8b^3 - 54a^2b + 36ab^2$

We use the algebraic identity: $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$.
Here,$a = \frac{1}{x}$ and $b = \frac{y}{3}$.
Substituting these values into the identity:
$\left(\frac{1}{x} + \frac{y}{3}\right)^{3} = \left(\frac{1}{x}\right)^{3} + \left(\frac{y}{3}\right)^{3} + 3 \left(\frac{1}{x}\right) \left(\frac{y}{3}\right) \left(\frac{1}{x} + \frac{y}{3}\right)$
$= \frac{1}{x^{3}} + \frac{y^{3}}{27} + \frac{y}{x} \left(\frac{1}{x} + \frac{y}{3}\right)$
$= \frac{1}{x^{3}} + \frac{y^{3}}{27} + \frac{y}{x^{2}} + \frac{y^{2}}{3x}$
Rearranging the terms,we get: $\frac{1}{x^{3}} + \frac{y}{x^{2}} + \frac{y^{2}}{3x} + \frac{y^{3}}{27}$.

(D) We use the algebraic identity $(a-b)^{3} = a^{3} - b^{3} - 3ab(a-b)$.
Here,$a = 4$ and $b = \frac{1}{3x}$.
Substituting these values into the identity:
$\left(4-\frac{1}{3 x}\right)^{3} = (4)^{3} - \left(\frac{1}{3 x}\right)^{3} - 3(4)\left(\frac{1}{3 x}\right)\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{12}{3 x}\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{4}{x}\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{16}{x} + \frac{4}{3 x^{2}}$

(A) The given expression is $1-64 a^{3}-12 a+48 a^{2}$.
We can rewrite this expression as $1^{3}-(4 a)^{3}-3(1)(4 a)(1-4 a)$.
This matches the algebraic identity $(x-y)^{3} = x^{3}-y^{3}-3xy(x-y)$,where $x=1$ and $y=4a$.
Therefore,the expression simplifies to $(1-4 a)^{3}$.
Thus,the factorised form is $(1-4 a)(1-4 a)(1-4 a)$.

(A) The given expression is $8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$.
We can rewrite this expression in the form of the algebraic identity $(a+b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$.
Here,$a = 2p$ and $b = \frac{1}{5}$.
Substituting these values into the identity:
$(2p)^{3} + 3(2p)^{2}(\frac{1}{5}) + 3(2p)(\frac{1}{5})^{2} + (\frac{1}{5})^{3}$
$= 8p^{3} + 3(4p^{2})(\frac{1}{5}) + 3(2p)(\frac{1}{25}) + \frac{1}{125}$
$= 8p^{3} + \frac{12}{5}p^{2} + \frac{6}{25}p + \frac{1}{125}$.
Thus,the expression is equivalent to $(2p + \frac{1}{5})^{3}$,which can be written as $(2p + \frac{1}{5})(2p + \frac{1}{5})(2p + \frac{1}{5})$.

(N/A) The given expression is of the form $(a+b)(a^2-ab+b^2)$,where $a = \frac{x}{2}$ and $b = 2y$.
Using the algebraic identity $(a+b)(a^2-ab+b^2) = a^3+b^3$:
$\left(\frac{x}{2}+2 y\right)\left\{\left(\frac{x}{2}\right)^{2}-\left(\frac{x}{2}\right)(2 y)+(2 y)^{2}\right\} = \left(\frac{x}{2}\right)^{3}+(2 y)^{3}$
$= \frac{x^3}{8} + 8y^3$

$(X^6-1)$ We use the algebraic identity $(a-b)(a^{2}+ab+b^{2}) = a^{3}-b^{3}$.
Given expression: $(x^{2}-1)(x^{4}+x^{2}+1)$
Rewrite the expression as: $(x^{2}-1)((x^{2})^{2} + (x^{2})(1) + (1)^{2})$
Here,$a = x^{2}$ and $b = 1$.
Applying the identity: $(x^{2})^{3} - (1)^{3}$
$= x^{6} - 1$

(A) We have,
$1+64 x^{3} = (1)^{3} + (4 x)^{3}$
Using the algebraic identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$,where $a = 1$ and $b = 4x$:
$= (1 + 4x)((1)^{2} - (1)(4x) + (4x)^{2})$
$= (1 + 4x)(1 - 4x + 16x^{2})$
$= (1 + 4x)(16x^{2} - 4x + 1)$

(A) We use the algebraic identity: $a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$.
Given expression: $a^{3} - 2\sqrt{2}b^{3}$.
Rewrite the expression as: $a^{3} - (\sqrt{2}b)^{3}$.
Applying the identity where $x = a$ and $y = \sqrt{2}b$:
$= (a - \sqrt{2}b)((a)^{2} + (a)(\sqrt{2}b) + (\sqrt{2}b)^{2})$.
$= (a - \sqrt{2}b)(a^{2} + \sqrt{2}ab + 2b^{2})$.

(D) We use the algebraic identity: $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) = a^{3}+b^{3}+c^{3}-3abc$.
Given expression: $(2x - y + 3z)((2x)^{2} + (-y)^{2} + (3z)^{2} - (2x)(-y) - (-y)(3z) - (3z)(2x))$.
Here,$a = 2x$,$b = -y$,and $c = 3z$.
Applying the identity:
$= (2x)^{3} + (-y)^{3} + (3z)^{3} - 3(2x)(-y)(3z)$
$= 8x^{3} - y^{3} + 27z^{3} - 3(2x)(-y)(3z)$
$= 8x^{3} - y^{3} + 27z^{3} + 18xyz$.

(A) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.
Given expression: $a^{3}-8b^{3}-64c^{3}-24abc$
This can be rewritten as: $a^{3}+(-2b)^{3}+(-4c)^{3}-3(a)(-2b)(-4c)$.
Here,$x=a$,$y=-2b$,and $z=-4c$.
Applying the identity:
$= (a-2b-4c)((a)^{2}+(-2b)^{2}+(-4c)^{2}-(a)(-2b)-(-2b)(-4c)-(-4c)(a))$
$= (a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ca)$.

(A) We use the algebraic identity: $x^{3} + y^{3} + z^{3} - 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} - xy - yz - zx)$.
Given expression: $2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c$
This can be rewritten as: $(\sqrt{2} a)^{3} + (2 b)^{3} + (-3 c)^{3} - 3(\sqrt{2} a)(2 b)(-3 c)$.
Here,$x = \sqrt{2} a$,$y = 2 b$,and $z = -3 c$.
Applying the identity:
$= (\sqrt{2} a + 2 b - 3 c)((\sqrt{2} a)^{2} + (2 b)^{2} + (-3 c)^{2} - (\sqrt{2} a)(2 b) - (2 b)(-3 c) - (-3 c)(\sqrt{2} a))$
$= (\sqrt{2} a + 2 b - 3 c)(2 a^{2} + 4 b^{2} + 9 c^{2} - 2 \sqrt{2} a b + 6 b c + 3 \sqrt{2} c a)$.

(C) Let $a = \frac{1}{2}$,$b = \frac{1}{3}$,and $c = -\frac{5}{6}$.
First,calculate the sum $a + b + c$:
$a + b + c = \frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3 + 2 - 5}{6} = \frac{0}{6} = 0$.
We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.
Applying this identity to the given expression:
$\left(\frac{1}{2}\right)^{3} + \left(\frac{1}{3}\right)^{3} + \left(-\frac{5}{6}\right)^{3} = 3 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(-\frac{5}{6}\right)$.
$= 3 \times \frac{1}{6} \times \left(-\frac{5}{6}\right) = \frac{1}{2} \times \left(-\frac{5}{6}\right) = -\frac{5}{12}$.

(D) We have the expression: $(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$.
This can be rewritten as: $(0.2)^{3} + (-0.3)^{3} + (0.1)^{3}$.
Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.
Now,calculate the sum $a + b + c$:
$a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.
We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.
Substituting the values of $a$,$b$,and $c$:
$a^{3} + b^{3} + c^{3} = 3(0.2)(-0.3)(0.1)$.
Calculating the product:
$3 \times 0.2 = 0.6$
$0.6 \times (-0.3) = -0.18$
$-0.18 \times 0.1 = -0.018$.
Therefore,$(0.2)^{3} - (0.3)^{3} + (0.1)^{3} = -0.018$.

(D) Let $a = x-2y$,$b = 2y-3z$,and $c = 3z-x$.
Then,$a+b+c = (x-2y) + (2y-3z) + (3z-x) = 0$.
We know that if $a+b+c = 0$,then $a^3 + b^3 + c^3 = 3abc$.
Substituting the values of $a, b,$ and $c$ back into the identity,we get:
$(x-2y)^3 + (2y-3z)^3 + (3z-x)^3 = 3(x-2y)(2y-3z)(3z-x)$.

(B) Given expression: $x^{3}+y^{3}-12xy+64$.
We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
Here,let $a=x$,$b=y$,and $c=4$.
Then the expression becomes $x^{3}+y^{3}+4^{3}-3(x)(y)(4) = x^{3}+y^{3}+64-12xy$.
Substituting $a+b+c = x+y+4$ into the identity:
Since $x+y = -4$,we have $x+y+4 = -4+4 = 0$.
Therefore,$x^{3}+y^{3}-12xy+64 = (0)(x^{2}+y^{2}+16-xy-4y-4x) = 0$.

(C) Given expression is $x^{3}-8 y^{3}-36 x y-216$.
We can rewrite this as $x^{3} + (-2y)^{3} + (-6)^{3} - 3(x)(-2y)(-6)$.
This is in the form of the algebraic identity $a^{3} + b^{3} + c^{3} - 3abc = (a+b+c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$,where $a=x$,$b=-2y$,and $c=-6$.
Substituting these values,we get: $(x - 2y - 6)(x^{2} + 4y^{2} + 36 + 2xy - 12y + 6x)$.
Since we are given $x = 2y + 6$,it follows that $x - 2y - 6 = 0$.
Therefore,the entire expression becomes $0 \times (x^{2} + 4y^{2} + 36 + 2xy - 12y + 6x) = 0$.

(A) Area of the rectangle is given by the polynomial $4 a^{2}+4 a-3$.
To find the length and breadth,we factorize the quadratic expression by splitting the middle term.
We need two numbers whose sum is $+4$ and whose product is $4 \times (-3) = -12$.
These two numbers are $+6$ and $-2$,since $6 + (-2) = 4$ and $6 \times (-2) = -12$.
Splitting the middle term $4 a$ as $6 a - 2 a$,we get:
$4 a^{2} + 6 a - 2 a - 3$
Grouping the terms:
$= 2 a(2 a + 3) - 1(2 a + 3)$
Factoring out the common binomial $(2 a + 3)$:
$= (2 a - 1)(2 a + 3)$
Since the area of a rectangle is defined as $\text{length} \times \text{breadth}$,the possible expressions for the dimensions are:
Length $= (2 a - 1)$ and Breadth $= (2 a + 3)$ or Length $= (2 a + 3)$ and Breadth $= (2 a - 1)$.

(A) We use the algebraic identity: $x^{3}+y^{3} = (x+y)^{3} - 3xy(x+y)$.
Given $x+y = 12$ and $xy = 27$.
Substituting these values into the identity:
$x^{3}+y^{3} = (12)^{3} - 3(27)(12)$.
Factor out $12$ to simplify the calculation:
$x^{3}+y^{3} = 12(12^{2} - 3 \times 27)$.
$x^{3}+y^{3} = 12(144 - 81)$.
$x^{3}+y^{3} = 12(63)$.
$x^{3}+y^{3} = 756$.

(B) Let $p(z) = az^{3} + 4z^{2} + 3z - 4$ and $q(z) = z^{3} - 4z + a$.
According to the Remainder Theorem,when a polynomial $f(z)$ is divided by $(z - c)$,the remainder is $f(c)$.
Since both polynomials leave the same remainder when divided by $z - 3$,we have $p(3) = q(3)$.
First,calculate $p(3)$:
$p(3) = a(3)^{3} + 4(3)^{2} + 3(3) - 4$
$p(3) = 27a + 36 + 9 - 4 = 27a + 41$.
Next,calculate $q(3)$:
$q(3) = (3)^{3} - 4(3) + a$
$q(3) = 27 - 12 + a = 15 + a$.
Equating the two remainders:
$27a + 41 = 15 + a$
$27a - a = 15 - 41$
$26a = -26$
$a = -1$.
Thus,the value of $a$ is $-1$.

(C) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - c)$,the remainder is $p(c)$.
Given $p(x) = x^{4} - 2x^{3} + 3x^{2} - ax + 3a - 7$ is divided by $x + 1$,the remainder is $p(-1) = 19$.
Substituting $x = -1$ into $p(x)$:
$p(-1) = (-1)^{4} - 2(-1)^{3} + 3(-1)^{2} - a(-1) + 3a - 7 = 19$
$1 - 2(-1) + 3(1) + a + 3a - 7 = 19$
$1 + 2 + 3 + 4a - 7 = 19$
$4a - 1 = 19$
$4a = 20$
$a = 5$
Now,we find the remainder when $p(x)$ is divided by $x + 2$,which is $p(-2)$ with $a = 5$:
$p(x) = x^{4} - 2x^{3} + 3x^{2} - 5x + 3(5) - 7 = x^{4} - 2x^{3} + 3x^{2} - 5x + 8$
$p(-2) = (-2)^{4} - 2(-2)^{3} + 3(-2)^{2} - 5(-2) + 8$
$p(-2) = 16 - 2(-8) + 3(4) + 10 + 8$
$p(-2) = 16 + 16 + 12 + 10 + 8 = 62$.

(A) Let $f(x) = p x^{2} + 5 x + r$.
Since $(x-2)$ is a factor of $f(x)$,by the Factor Theorem,$f(2) = 0$.
$p(2)^{2} + 5(2) + r = 0$
$4p + 10 + r = 0$
$4p + r = -10$ --- $(1)$
Since $(x - \frac{1}{2})$ is a factor of $f(x)$,by the Factor Theorem,$f(\frac{1}{2}) = 0$.
$p(\frac{1}{2})^{2} + 5(\frac{1}{2}) + r = 0$
$\frac{1}{4}p + \frac{5}{2} + r = 0$
Multiply the entire equation by $4$:
$p + 10 + 4r = 0$
$p + 4r = -10$ --- $(2)$
Comparing equations $(1)$ and $(2)$,since both are equal to $-10$:
$4p + r = p + 4r$
$4p - p = 4r - r$
$3p = 3r$
$p = r$
Hence,it is proved.

(N/A) Let $p(x) = 2x^4 - 5x^3 + 2x^2 - x + 2$ and $g(x) = x^2 - 3x + 2$.
First,factorize the divisor $g(x)$:
$g(x) = x^2 - 3x + 2 = x^2 - x - 2x + 2 = x(x - 1) - 2(x - 1) = (x - 1)(x - 2)$.
For $p(x)$ to be divisible by $g(x)$,it must be divisible by both $(x - 1)$ and $(x - 2)$ by the Factor Theorem.
Check for $(x - 1)$:
$p(1) = 2(1)^4 - 5(1)^3 + 2(1)^2 - 1 + 2 = 2 - 5 + 2 - 1 + 2 = 0$.
Since $p(1) = 0$,$(x - 1)$ is a factor of $p(x)$.
Check for $(x - 2)$:
$p(2) = 2(2)^4 - 5(2)^3 + 2(2)^2 - 2 + 2 = 2(16) - 5(8) + 2(4) - 2 + 2 = 32 - 40 + 8 = 0$.
Since $p(2) = 0$,$(x - 2)$ is a factor of $p(x)$.
Since both $(x - 1)$ and $(x - 2)$ are factors of $p(x)$ and they are coprime,their product $(x - 1)(x - 2) = x^2 - 3x + 2$ must also be a factor of $p(x)$.
Hence,$2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2$.

(A) We use the algebraic identity $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$.
Let $a = (2x - 5y)$ and $b = (2x + 5y)$.
Then,$a - b = (2x - 5y) - (2x + 5y) = 2x - 5y - 2x - 5y = -10y$.
Now,calculate $a^{2} + ab + b^{2}$:
$a^{2} = (2x - 5y)^{2} = 4x^{2} + 25y^{2} - 20xy$
$b^{2} = (2x + 5y)^{2} = 4x^{2} + 25y^{2} + 20xy$
$ab = (2x - 5y)(2x + 5y) = 4x^{2} - 25y^{2}$
Summing these: $(4x^{2} + 25y^{2} - 20xy) + (4x^{2} - 25y^{2}) + (4x^{2} + 25y^{2} + 20xy) = 12x^{2} + 25y^{2}$.
Finally,multiply $(a - b)$ and $(a^{2} + ab + b^{2})$:
$(-10y)(12x^{2} + 25y^{2}) = -120x^{2}y - 250y^{3}$.

(D) We have the expression: $(-z+x-2 y)(x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z)$.
Rearranging the terms,we get: $(x-2 y-z)(x^{2}+(-2 y)^{2}+(-z)^{2}-(x)(-2 y)-(-2 y)(-z)-(x)(-z))$.
This expression is in the form $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$,where $a=x$,$b=-2 y$,and $c=-z$.
Using the algebraic identity $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) = a^{3}+b^{3}+c^{3}-3abc$,we substitute the values:
$= (x)^{3} + (-2 y)^{3} + (-z)^{3} - 3(x)(-2 y)(-z)$.
$= x^{3} - 8 y^{3} - z^{3} - 6 x y z$.

(N/A) Given that $a, b, c$ are non-zero and $a+b+c=0$.
We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
Since $a+b+c=0$,the right side becomes $0$,so $a^{3}+b^{3}+c^{3}-3abc = 0$,which implies $a^{3}+b^{3}+c^{3}=3abc$.
Now,consider the expression $\frac{a^{2}}{bc} + \frac{b^{2}}{ca} + \frac{c^{2}}{ab}$.
Taking the least common multiple $(LCM)$ of the denominators $bc, ca, ab$,which is $abc$,we get:
$\frac{a^{2}(a) + b^{2}(b) + c^{2}(c)}{abc} = \frac{a^{3}+b^{3}+c^{3}}{abc}$.
Substituting $a^{3}+b^{3}+c^{3} = 3abc$ into the expression:
$\frac{3abc}{abc} = 3$.
Thus,the expression equals $3$.

(N/A) We know the algebraic identity:
$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$= (a+b+c)[a^2+b^2+c^2-(ab+bc+ca)]$
Given $a+b+c=5$ and $ab+bc+ca=10$,we substitute these values:
$= 5[a^2+b^2+c^2-10]$
Now,we find $a^2+b^2+c^2$ using the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$:
$(5)^2 = a^2+b^2+c^2+2(10)$
$25 = a^2+b^2+c^2+20$
$a^2+b^2+c^2 = 25-20 = 5$
Substituting this value back into the expression:
$a^3+b^3+c^3-3abc = 5(5-10) = 5(-5) = -25$.
Hence,it is proved.

(N/A) We know the expansion of $(a+b+c)^3$ is given by:
$(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)$
Alternatively,we can expand it step-by-step:
$(a+b+c)^3 = [(a+b)+c]^3$
$= (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3$
$= (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + b^2)c + 3ac^2 + 3bc^2 + c^3$
$= a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2$
Now,subtract $(a^3 + b^3 + c^3)$ from both sides:
$(a+b+c)^3 - a^3 - b^3 - c^3 = 3a^2b + 3ab^2 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2$
Factor out $3$ from the right side:
$= 3[a^2b + ab^2 + a^2c + 2abc + b^2c + ac^2 + bc^2]$
$= 3[ab(a+b) + ac(a+c) + bc(b+c) + 2abc]$
$= 3[ab(a+b) + ac(a+c) + bc(b+c) + abc + abc]$
$= 3[ab(a+b) + abc + ac(a+c) + ac^2 + bc(b+c) + abc]$
$= 3[ab(a+b+c) + ac(a+c+b) + bc(b+c+a)]$
$= 3(a+b+c)(ab + ac + bc)$
Wait,the standard identity is $(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a)$.
Let's verify: $3(a+b)(b+c)(c+a) = 3(ab + ac + b^2 + bc)(c+a)$
$= 3(abc + a^2b + ac^2 + a^2c + b^2c + ab^2 + bc^2 + abc)$
$= 3(a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc)$.
This matches the expansion derived above.
Hence,$(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a)$ is proved.

(A) The given expression is $\pi x^{2}-\sqrt{3} x+11$.
$1$. $A$ polynomial is an algebraic expression in which the exponents of the variable are non-negative integers.
$2$. In the expression $\pi x^{2}-\sqrt{3} x+11$,the exponents of the variable $x$ are $2$,$1$,and $0$ (since $11 = 11x^{0}$).
$3$. Since all exponents are non-negative integers,the expression is a polynomial.
$4$. Since the expression contains only one variable,$x$,it is a polynomial in one variable.

(D) The given expression is $x^{2} + x - 3 + \frac{4}{x}$.
We can rewrite the term $\frac{4}{x}$ as $4x^{-1}$.
Thus,the expression becomes $x^{2} + x - 3 + 4x^{-1}$.
$A$ polynomial is defined as an algebraic expression where the exponents of the variables are non-negative integers.
In this expression,the exponent of $x$ in the term $4x^{-1}$ is $-1$,which is a negative integer.
Therefore,the given expression is not a polynomial.

(D) The given expression is $5 x^{2}-7 x+3 \sqrt{x}$.
To determine if an expression is a polynomial,the exponent of the variable must be a non-negative integer.
In the given expression,the term $3 \sqrt{x}$ can be written as $3 x^{\frac{1}{2}}$.
Here,the exponent of $x$ is $\frac{1}{2}$,which is not a non-negative integer.
Therefore,$5 x^{2}-7 x+3 \sqrt{x}$ is not a polynomial.

(N/A) The given expression $x^{3}+y^{3}+z^{3}-3xyz$ is a polynomial because the exponent of the variable in each of its terms is a non-negative integer (a whole number).
However,the expression contains three distinct variables: $x$,$y$,and $z$.
Therefore,it is a polynomial,but it is not a polynomial in one variable.