Simplification Questions in English

(D) Given the equation: $\sqrt{1+\frac{x}{144}} = \frac{13}{12}$
Square both sides of the equation to remove the square root:
$1 + \frac{x}{144} = \left(\frac{13}{12}\right)^2$
$1 + \frac{x}{144} = \frac{169}{144}$
Subtract $1$ from both sides:
$\frac{x}{144} = \frac{169}{144} - 1$
$\frac{x}{144} = \frac{169 - 144}{144}$
$\frac{x}{144} = \frac{25}{144}$
Multiplying both sides by $144$,we get:
$x = 25$

(B) Given: $a = \sqrt{2} + 1$ and $b = \sqrt{2} - 1.$
We need to find the value of $\frac{1}{a + 1} + \frac{1}{b + 1}.$
Substitute the values of $a$ and $b$:
$\frac{1}{(\sqrt{2} + 1) + 1} + \frac{1}{(\sqrt{2} - 1) + 1} = \frac{1}{\sqrt{2} + 2} + \frac{1}{\sqrt{2}}.$
To simplify,rationalize the first term:
$\frac{1}{\sqrt{2} + 2} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{2 - \sqrt{2}}{4 - 2} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}.$
Now add the second term:
$(1 - \frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = 1.$
Thus,the value is $1$.

(C) Given that $x + \frac{1}{x} = \sqrt{13}$.
Divide the numerator and denominator of the expression $\frac{3x}{x^2 - 1}$ by $x$:
$\frac{3x/x}{(x^2 - 1)/x} = \frac{3}{x - \frac{1}{x}}$.
We know that $(x - \frac{1}{x})^2 = (x + \frac{1}{x})^2 - 4$.
Substituting the given value: $(x - \frac{1}{x})^2 = (\sqrt{13})^2 - 4 = 13 - 4 = 9$.
Therefore,$x - \frac{1}{x} = \pm 3$.
Substituting this into our expression: $\frac{3}{\pm 3} = \pm 1$.
Since the options provided are limited,we consider the magnitude or the positive case,which is $1$.

(D) Given equation: $\sqrt{575} \div x \times 14.98^{2} = 450$
Step $1$: Approximate the values. $\sqrt{575} \approx \sqrt{576} = 24$ and $14.98 \approx 15$.
Step $2$: Substitute the approximated values into the equation: $24 \div x \times 15^{2} = 450$.
Step $3$: Simplify the equation: $\frac{24}{x} \times 225 = 450$.
Step $4$: Solve for $x$: $\frac{24}{x} = \frac{450}{225}$.
Step $5$: $\frac{24}{x} = 2$.
Step $6$: $x = \frac{24}{2} = 12$.
Therefore,the value is $12$.

(C) To solve the equation $30.01^{2} - 19.98^{2} - ? = 21.81^{2}$,we can approximate the values to the nearest integers:
$30.01 \approx 30$
$19.98 \approx 20$
$21.81 \approx 22$
Substituting these values into the equation:
$30^{2} - 20^{2} - x = 22^{2}$
Calculate the squares:
$900 - 400 - x = 484$
Simplify the left side:
$500 - x = 484$
Solve for $x$:
$x = 500 - 484$
$x = 16$

(B) To solve the expression $820.15 + 2379.85 + 140.01 \times 4.99$,we use the $BODMAS$ rule.
First,we round the numbers to their nearest integers for approximation:
$820.15 \approx 820$
$2379.85 \approx 2380$
$140.01 \approx 140$
$4.99 \approx 5$
Now,substitute these values into the expression:
$820 + 2380 + (140 \times 5)$
$= 3200 + 700$
$= 3900$
Therefore,the value that should come in place of the question mark is $3900$.

(C) To solve the equation,we approximate the values to the nearest integers:
$39.97 \% \approx 40 \% = 0.40$
$649.8 \approx 650$
$13.05 \approx 13$
$45.12 \approx 45$
Substituting these values into the equation:
$0.40 \times 650 \div 13 = 45 - ?$
$260 \div 13 = 45 - ?$
$20 = 45 - ?$
$? = 45 - 20 = 25$
Therefore,the value is $25$.

(D) To solve the expression $(674.87 + 59.98) \div 35.02$,we use approximation to simplify the calculation.
$674.87 \approx 675$
$59.98 \approx 60$
$35.02 \approx 35$
Now,substitute these values into the expression:
$(675 + 60) \div 35$
$= 735 \div 35$
$= 21$
Therefore,the value that should come in place of the question mark is $21$.

(C) To find the approximate value,we round the numbers to the nearest whole numbers:
$\frac{1810}{24} \times 8 + 11 \times 19 = x - 306$
First,calculate the division and multiplication:
$\frac{1810}{24} \approx 75.416$
$75.416 \times 8 \approx 603.33$
$11 \times 19 = 209$
Now,substitute these values into the equation:
$603 + 209 = x - 306$
$812 = x - 306$
Solve for $x$:
$x = 812 + 306$
$x = 1118$

(A) Given equation: $2775 \times \frac{160}{\sqrt{x}} = 5550$
Rearranging the equation to solve for $\sqrt{x}$:
$\sqrt{x} = \frac{2775 \times 160}{5550}$
Since $2775 \times 2 = 5550$,we can simplify the fraction:
$\sqrt{x} = \frac{160}{2}$
$\sqrt{x} = 80$
Squaring both sides:
$x = 80^2 = 6400$

(D) To find the approximate value,we round the numbers to the nearest integers:
$24.98 \approx 25$,$16.02 \approx 16$,$7.98 \approx 8$,$15.04 \approx 15$,$38.93 \approx 39$.
Substituting these values into the equation:
$25^{2} \times \frac{16^{2}}{8 \times 15} \times 39 = 130 \times x^{2}$
$\frac{625 \times 256}{120} \times 39 = 130 \times x^{2}$
Simplify the expression:
$\frac{625 \times 256 \times 39}{120} = 130 \times x^{2}$
$x^{2} = \frac{625 \times 256 \times 39}{120 \times 130}$
$x^{2} = \frac{625 \times 256 \times 39}{15600} = \frac{6240000}{15600} = 400$
$x = \sqrt{400} = 20$.

(D) To find the approximate value,we round the percentages and numbers:
$71.98 \% \approx 72 \%$
$35.06 \% \approx 35 \%$
The equation becomes:
$72 \% \text{ of } 1200 + 35 \% \text{ of } 270 = x \% \text{ of } 600$
Calculate the values:
$72 \% \text{ of } 1200 = 0.72 \times 1200 = 864$
$35 \% \text{ of } 270 = 0.35 \times 270 = 94.5$
Summing these values:
$864 + 94.5 = 958.5$
Now,solve for $x$:
$958.5 = \frac{x}{100} \times 600$
$958.5 = 6x$
$x = \frac{958.5}{6} = 159.75$
Rounding to the nearest whole number,we get $x \approx 160$.

(A) To find the approximate value,we round the numbers to the nearest whole numbers:
$\frac{7702}{44} + 25 \times 46 = x \times 15$
First,calculate the division:
$7702 \div 44 \approx 175.045 \approx 175$
Next,calculate the multiplication:
$25 \times 46 = 1150$
Now,add the results:
$175 + 1150 = 1325$
Finally,solve for $x$:
$1325 = x \times 15$
$x = \frac{1325}{15} \approx 88.33$
Rounding to the nearest integer,we get $x \approx 88$.

(D) To compare the values,we use the approximate values of the square roots:
$\sqrt{2} \approx 1.414$
$\sqrt{3} \approx 1.732$
$\sqrt{5} \approx 2.236$
$\sqrt{6} \approx 2.449$
Now,calculate the sums:
$\sqrt{6} + \sqrt{2} \approx 2.449 + 1.414 = 3.863$
$\sqrt{5} + \sqrt{3} \approx 2.236 + 1.732 = 3.968$
Comparing the two sums:
$3.863 < 3.968$
Therefore,$\sqrt{6} + \sqrt{2} < \sqrt{5} + \sqrt{3}$.
This means statement $(ii)$ is correct,while statements $(i)$ and $(iii)$ are incorrect.

(B) Let $a = 0.67$ and $b = 0.33$.
The given expression is in the form $\frac{a^3 - b^3}{a^2 + ab + b^2}$.
Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,we can simplify the expression:
$\frac{(a - b)(a^2 + ab + b^2)}{a^2 + ab + b^2} = a - b$.
Substituting the values of $a$ and $b$ back into the simplified expression:
$a - b = 0.67 - 0.33 = 0.34$.

(B) First,simplify the denominator:
$\sqrt{7+4 \sqrt{3}} = \sqrt{7+2 \times 2 \times \sqrt{3}} = \sqrt{4+3+2 \times 2 \times \sqrt{3}} = \sqrt{(2+\sqrt{3})^{2}} = 2+\sqrt{3}$
$\sqrt{4+2 \sqrt{3}} = \sqrt{3+1+2 \times \sqrt{3} \times 1} = \sqrt{(\sqrt{3}+1)^{2}} = \sqrt{3}+1$
Denominator $= (2+\sqrt{3}) - (\sqrt{3}+1) = 2+\sqrt{3}-\sqrt{3}-1 = 1$
Next,calculate the numerator using the sum of squares formula $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$:
Sum of squares from $1$ to $10 = \frac{10(11)(21)}{6} = 385$
Sum of squares from $1$ to $5 = \frac{5(6)(11)}{6} = 55$
Numerator $= 6^{2}+7^{2}+8^{2}+9^{2}+10^{2} = 385 - 55 = 330$
Result $= \frac{330}{1} = 330$

(D) Given that $\sqrt[3]{79507} = 43$.
We need to evaluate the expression: $\sqrt[3]{79.507} + \sqrt[3]{0.079507} + \sqrt[3]{0.000079507}$.
$1$. $\sqrt[3]{79.507} = \sqrt[3]{\frac{79507}{1000}} = \frac{43}{10} = 4.3$.
$2$. $\sqrt[3]{0.079507} = \sqrt[3]{\frac{79507}{1000000}} = \frac{43}{100} = 0.43$.
$3$. $\sqrt[3]{0.000079507} = \sqrt[3]{\frac{79507}{1000000000}} = \frac{43}{1000} = 0.043$.
Adding these values together:
$4.3 + 0.43 + 0.043 = 4.773$.

(A) Given expression: $\frac{(0.064-0.008)(0.16-0.04)}{(0.16+0.08+0.04)(0.4+0.2)^{3}}$
Express the terms as powers of $0.4$ and $0.2$:
$0.064 = 0.4^{3}$,$0.008 = 0.2^{3}$,$0.16 = 0.4^{2}$,$0.04 = 0.2^{2}$,$0.08 = 0.4 \times 0.2$
Substitute these into the expression:
Numerator: $(0.4^{3}-0.2^{3})(0.4^{2}-0.2^{2})$
Denominator: $(0.4^{2}+0.4 \times 0.2+0.2^{2})(0.4+0.2)^{3}$
Using algebraic identities $a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})$ and $a^{2}-b^{2} = (a-b)(a+b)$:
Numerator: $(0.4-0.2)(0.4^{2}+0.4 \times 0.2+0.2^{2}) \times (0.4-0.2)(0.4+0.2)$
Now the expression becomes:
$\frac{(0.4-0.2)(0.4^{2}+0.4 \times 0.2+0.2^{2})(0.4-0.2)(0.4+0.2)}{(0.4^{2}+0.4 \times 0.2+0.2^{2})(0.4+0.2)^{3}}$
Cancel the common terms $(0.4^{2}+0.4 \times 0.2+0.2^{2})$ and one $(0.4+0.2)$:
$= \frac{(0.4-0.2)^{2}}{(0.4+0.2)^{2}} = \frac{(0.2)^{2}}{(0.6)^{2}} = \frac{0.04}{0.36} = \frac{4}{36} = \frac{1}{9}$

(A) Given the expression $x = \sqrt{a \sqrt[3]{b \sqrt{a \sqrt[3]{b \dots \infty}}}}.$
Squaring both sides,we get $x^2 = a \sqrt[3]{b \sqrt{a \sqrt[3]{b \dots \infty}}}.$
Now,cubing both sides of this equation,we get $(x^2)^3 = a^3 \cdot b \sqrt{a \sqrt[3]{b \dots \infty}}.$
This simplifies to $x^6 = a^3 b x.$
Since $x \neq 0,$ we can divide by $x$ to get $x^5 = a^3 b.$
Therefore,$x = \sqrt[5]{a^3 b}.$

(B) Given that $\frac{x}{y} = \frac{3}{4}.$ Let $x = 3k$ and $y = 4k$ for some constant $k \neq 0.$
We need to find the ratio $\frac{2x + 3y}{3y - 2x}.$
Substituting the values of $x$ and $y$:
$\frac{2(3k) + 3(4k)}{3(4k) - 2(3k)} = \frac{6k + 12k}{12k - 6k}$
$= \frac{18k}{6k} = \frac{18}{6} = \frac{3}{1}.$
Therefore,the ratio is $3:1$.

(B) To solve the expression,first remove the decimals by adjusting the powers of $10$ in the numerator and denominator.
$\sqrt{\frac{0.324 \times 0.081 \times 4.624}{1.5625 \times 0.0289 \times 72.9 \times 64}} = \sqrt{\frac{324 \times 10^{-3} \times 81 \times 10^{-3} \times 4624 \times 10^{-3}}{15625 \times 10^{-4} \times 289 \times 10^{-4} \times 729 \times 10^{-1} \times 64}}$
$= \sqrt{\frac{324 \times 81 \times 4624 \times 10^{-9}}{15625 \times 289 \times 729 \times 64 \times 10^{-9}}}$
$= \sqrt{\frac{324 \times 81 \times 4624}{15625 \times 289 \times 729 \times 64}}$
$= \sqrt{\frac{18^2 \times 9^2 \times 68^2}{125^2 \times 17^2 \times 27^2 \times 8^2}}$
$= \frac{18 \times 9 \times 68}{125 \times 17 \times 27 \times 8}$
$= \frac{11016}{459000} = 0.024$

(D) Given: $\frac{x^{24}+1}{x^{12}}=7$
Dividing each term in the numerator by the denominator:
$\frac{x^{24}}{x^{12}}+\frac{1}{x^{12}}=7$
$x^{12}+\frac{1}{x^{12}}=7$ ..... $(1)$
We need to find the value of $\frac{x^{72}+1}{x^{36}}$,which is $x^{36}+\frac{1}{x^{36}}$.
Using the identity $(a+b)^3 = a^3+b^3+3ab(a+b)$,let $a=x^{12}$ and $b=\frac{1}{x^{12}}$:
$\left(x^{12}+\frac{1}{x^{12}}\right)^3 = (x^{12})^3 + \left(\frac{1}{x^{12}}\right)^3 + 3(x^{12})\left(\frac{1}{x^{12}}\right)\left(x^{12}+\frac{1}{x^{12}}\right)$
Substitute the value from equation $(1)$:
$7^3 = x^{36} + \frac{1}{x^{36}} + 3(1)(7)$
$343 = x^{36} + \frac{1}{x^{36}} + 21$
$x^{36} + \frac{1}{x^{36}} = 343 - 21$
$x^{36} + \frac{1}{x^{36}} = 322$
Thus,$\frac{x^{72}+1}{x^{36}} = 322$.

(A) Given expression: $x^{3}+27x^{2}+243x+631$.
Substitute $x=2$ into the expression:
$= (2)^{3} + 27(2)^{2} + 243(2) + 631$
$= 8 + 27(4) + 486 + 631$
$= 8 + 108 + 486 + 631$
$= 1233$.

(A) Given expression: $p(p^{2} + 3p + 3) = p^{3} + 3p^{2} + 3p$.
To simplify this,we can add and subtract $1$ to complete the cube formula: $(p+1)^{3} = p^{3} + 3p^{2} + 3p + 1$.
Therefore,$p^{3} + 3p^{2} + 3p = (p+1)^{3} - 1$.
Substitute $p = 99$ into the expression:
$(99 + 1)^{3} - 1 = 100^{3} - 1$.
$100^{3} = 1000000$.
$1000000 - 1 = 999999$.

(B) Given the equation: $a^{2}+b^{2}+c^{2}=ab+bc+ca$.
Rearranging the terms,we get: $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$.
Multiplying the entire equation by $2$,we get: $2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca=0$.
This can be rewritten as: $(a^{2}-2ab+b^{2})+(b^{2}-2bc+c^{2})+(c^{2}-2ca+a^{2})=0$.
This simplifies to: $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$.
Since the sum of squares is zero,each term must individually be zero:
$a-b=0 \Rightarrow a=b$,
$b-c=0 \Rightarrow b=c$,
$c-a=0 \Rightarrow c=a$.
Thus,$a=b=c$.
Substituting these values into the expression $\frac{a+c}{b}$:
$\frac{a+c}{b} = \frac{a+a}{a} = \frac{2a}{a} = 2$.

(A) Given $ab + bc + ca = 0$.
From this,we have $bc = -ab - ca$.
Substituting this into the first term denominator: $a^2 - bc = a^2 - (-ab - ca) = a^2 + ab + ca = a(a + b + c)$.
Similarly,for the other denominators:
$b^2 - ca = b^2 - (-ab - bc) = b^2 + ab + bc = b(a + b + c)$.
$c^2 - ab = c^2 - (-bc - ca) = c^2 + bc + ca = c(a + b + c)$.
Now,substitute these into the expression:
$\frac{1}{a(a + b + c)} + \frac{1}{b(a + b + c)} + \frac{1}{c(a + b + c)}$
$= \frac{1}{a + b + c} \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$
$= \frac{1}{a + b + c} \left(\frac{bc + ac + ab}{abc}\right)$
Since $ab + bc + ca = 0$,the numerator of the second fraction is $0$.
$= \frac{1}{a + b + c} \times \frac{0}{abc} = 0$.

(D) Given that $(2+\sqrt{3}) a = 1$ and $(2-\sqrt{3}) b = 1$.
From $(2+\sqrt{3}) a = 1$,we get $a = \frac{1}{2+\sqrt{3}}$.
Therefore,$\frac{1}{a} = 2+\sqrt{3}$.
Similarly,from $(2-\sqrt{3}) b = 1$,we get $b = \frac{1}{2-\sqrt{3}}$.
Therefore,$\frac{1}{b} = 2-\sqrt{3}$.
Now,adding the two values:
$\frac{1}{a} + \frac{1}{b} = (2+\sqrt{3}) + (2-\sqrt{3}) = 4$.

(B) Given the equation: $3x + \frac{3}{x} = 1$.
Divide the entire equation by $3$ to get: $x + \frac{1}{x} = \frac{1}{3}$.
Now,cube both sides of the equation using the identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$:
$(x + \frac{1}{x})^{3} = (\frac{1}{3})^{3}$
$x^{3} + \frac{1}{x^{3}} + 3(x)(\frac{1}{x})(x + \frac{1}{x}) = \frac{1}{27}$
Substitute the value $x + \frac{1}{x} = \frac{1}{3}$ into the equation:
$x^{3} + \frac{1}{x^{3}} + 3(1)(\frac{1}{3}) = \frac{1}{27}$
$x^{3} + \frac{1}{x^{3}} + 1 = \frac{1}{27}$.

(D) Let the expression be $E = \frac{1}{a^{2}+a x+x^{2}}-\frac{1}{a^{2}-a x+x^{2}}+\frac{2 a x}{a^{4}+a^{2} x^{2}+x^{4}}$.
First,simplify the first two terms by taking the common denominator $(a^{2}+a x+x^{2})(a^{2}-a x+x^{2})$.
Note that $(a^{2}+a x+x^{2})(a^{2}-a x+x^{2}) = (a^{2}+x^{2})^{2} - (ax)^{2} = a^{4} + 2a^{2}x^{2} + x^{4} - a^{2}x^{2} = a^{4} + a^{2}x^{2} + x^{4}$.
So,the expression becomes:
$E = \frac{(a^{2}-a x+x^{2}) - (a^{2}+a x+x^{2})}{a^{4}+a^{2}x^{2}+x^{4}} + \frac{2 a x}{a^{4}+a^{2} x^{2}+x^{4}}$
$E = \frac{a^{2}-a x+x^{2}-a^{2}-a x-x^{2}}{a^{4}+a^{2}x^{2}+x^{4}} + \frac{2 a x}{a^{4}+a^{2} x^{2}+x^{4}}$
$E = \frac{-2ax}{a^{4}+a^{2}x^{2}+x^{4}} + \frac{2ax}{a^{4}+a^{2} x^{2}+x^{4}}$
$E = 0$.

(B) Let $a = 941$ and $b = 149$.
The given expression is of the form $\frac{(a+b)^{2} + (a-b)^{2}}{a^{2} + b^{2}}$.
Expanding the numerator using algebraic identities $(a+b)^{2} = a^{2} + b^{2} + 2ab$ and $(a-b)^{2} = a^{2} + b^{2} - 2ab$:
Numerator $= (a^{2} + b^{2} + 2ab) + (a^{2} + b^{2} - 2ab) = 2(a^{2} + b^{2})$.
Substituting this back into the expression:
$\frac{2(a^{2} + b^{2})}{a^{2} + b^{2}} = 2$.
Therefore,the value of the expression is $2$.

(A) Given the expression: $5 \sqrt{5} \times 5^{3} \div 5^{-\frac{3}{2}} = 5^{a+2}$
Express all terms with base $5$:
$5^1 \times 5^{1/2} \times 5^3 \div 5^{-3/2} = 5^{a+2}$
Using the laws of exponents $x^m \times x^n = x^{m+n}$ and $x^m \div x^n = x^{m-n}$:
$5^{(1 + 1/2 + 3 - (-3/2))} = 5^{a+2}$
Simplify the exponent:
$1 + 0.5 + 3 + 1.5 = 6$
So,$5^6 = 5^{a+2}$
Equating the exponents:
$6 = a + 2$
$a = 6 - 2 = 4$

(A) Let $x = 3+2 \sqrt{2}$ and $y = 3-2 \sqrt{2}$.
Note that $xy = (3+2 \sqrt{2})(3-2 \sqrt{2}) = 3^2 - (2 \sqrt{2})^2 = 9 - 8 = 1$.
We need to find $x^{-3} + y^{-3} = \frac{1}{x^3} + \frac{1}{y^3} = \frac{x^3 + y^3}{(xy)^3}$.
Since $xy = 1$,this simplifies to $x^3 + y^3$.
Using the identity $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$:
$x+y = (3+2 \sqrt{2}) + (3-2 \sqrt{2}) = 6$.
$x^3 + y^3 = (6)^3 - 3(1)(6) = 216 - 18 = 198$.

(A) Given expression: $\left\{\left(\sqrt[n]{x^{2}}\right)^{n / 2}\right\}^{2}$
Step $1$: Simplify the inner term $\sqrt[n]{x^{2}}$. This can be written as $(x^{2})^{1/n} = x^{2/n}$.
Step $2$: Substitute this back into the expression: $\left\{(x^{2/n})^{n/2}\right\}^{2}$.
Step $3$: Use the power rule $(a^{m})^{n} = a^{m \cdot n}$. The expression becomes $\left\{x^{(2/n) \cdot (n/2)}\right\}^{2}$.
Step $4$: Simplify the exponent: $(2/n) \cdot (n/2) = 1$. So,the expression simplifies to $\{x^{1}\}^{2}$.
Step $5$: Finally,$x^{1 \cdot 2} = x^{2}$.

(B) Given equation: $(\sqrt{3})^{5} \times 9^{2} = 3^{n} \times 3 \sqrt{3}$
Express all terms with base $3$:
$(\sqrt{3})^{5} = (3^{1/2})^{5} = 3^{5/2}$
$9^{2} = (3^{2})^{2} = 3^{4}$
$3 \sqrt{3} = 3^{1} \times 3^{1/2} = 3^{1 + 1/2} = 3^{3/2}$
Substitute these into the equation:
$3^{5/2} \times 3^{4} = 3^{n} \times 3^{3/2}$
Using the law of exponents $a^{m} \times a^{n} = a^{m+n}$:
$3^{5/2 + 4} = 3^{n + 3/2}$
$3^{13/2} = 3^{n + 3/2}$
Equating the exponents:
$13/2 = n + 3/2$
$n = 13/2 - 3/2$
$n = 10/2$
$n = 5$

(A) Given expression: $p(p^{2}+3p+3)$
$= p^{3}+3p^{2}+3p$
To complete the algebraic identity $(p+1)^{3} = p^{3}+3p^{2}+3p+1$,we add and subtract $1$:
$= (p^{3}+3p^{2}+3p+1) - 1$
$= (p+1)^{3} - 1$
Substitute $p=99$:
$= (99+1)^{3} - 1$
$= 100^{3} - 1$
$= 1000000 - 1$
$= 999999$

(B) Given equation: $3^{x}-3^{x-1}=486$
We can rewrite $3^{x-1}$ as $\frac{3^{x}}{3}$:
$3^{x}-\frac{3^{x}}{3}=486$
Factor out $3^{x}$:
$3^{x}(1-\frac{1}{3})=486$
Simplify the expression inside the parentheses:
$3^{x}(\frac{2}{3})=486$
Multiply both sides by $\frac{3}{2}$:
$3^{x}=486 \times \frac{3}{2}$
$3^{x}=243 \times 3$
$3^{x}=729$
Express $729$ as a power of $3$:
$729 = 3^{6}$
Therefore,$3^{x}=3^{6}$,which implies $x=6$.

(B) Let $a = 2.75$ and $b = 2.25$.
The given expression is of the form $\frac{a^3 - b^3}{a^2 + ab + b^2}$.
We know the algebraic identity: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.
Substituting this into the expression,we get:
$\frac{(a - b)(a^2 + ab + b^2)}{a^2 + ab + b^2} = a - b$.
Now,substitute the values of $a$ and $b$ back:
$2.75 - 2.25 = 0.5$.
Since $0.5 = \frac{1}{2}$,the correct option is $B$.

(B) To simplify the expression $1-\frac{a}{1-\frac{1}{1+\frac{a}{1-a}}}$,we start from the innermost fraction:
First,simplify the denominator $1+\frac{a}{1-a}$:
$1+\frac{a}{1-a} = \frac{1-a+a}{1-a} = \frac{1}{1-a}$
Next,substitute this back into the expression:
$1-\frac{a}{1-\frac{1}{\frac{1}{1-a}}} = 1-\frac{a}{1-(1-a)}$
Now,simplify the denominator $1-(1-a)$:
$1-1+a = a$
Finally,substitute this back:
$1-\frac{a}{a} = 1-1 = 0$
Thus,the correct option is $B$.

(D) Given expression: $\frac{(243)^{\frac{n}{5}} \times 3^{2n+1}}{9^{n} \times 3^{n-1}}$
Express all terms with base $3$:
$243 = 3^5$ and $9 = 3^2$.
Substitute these into the expression:
$\frac{(3^5)^{\frac{n}{5}} \times 3^{2n+1}}{(3^2)^n \times 3^{n-1}}$
Simplify the exponents:
$= \frac{3^n \times 3^{2n+1}}{3^{2n} \times 3^{n-1}}$
Combine the exponents using the rule $a^m \times a^n = a^{m+n}$:
$= \frac{3^{n + 2n + 1}}{3^{2n + n - 1}} = \frac{3^{3n+1}}{3^{3n-1}}$
Use the rule $\frac{a^m}{a^n} = a^{m-n}$:
$= 3^{(3n+1) - (3n-1)} = 3^{3n+1-3n+1} = 3^2 = 9$.

(C) The algebraic identity for $x^{3} + y^{3} + z^{3} - 3xyz$ is given by:
$x^{3} + y^{3} + z^{3} - 3xyz = \frac{1}{2}(x + y + z)[(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]$
Given $x = 333$,$y = 333$,and $z = 334$:
Sum $(x + y + z) = 333 + 333 + 334 = 1000$
Differences:
$(x - y) = 333 - 333 = 0$
$(y - z) = 333 - 334 = -1$
$(z - x) = 334 - 333 = 1$
Substituting these values into the identity:
$= \frac{1}{2}(1000)[(0)^{2} + (-1)^{2} + (1)^{2}]$
$= \frac{1}{2}(1000)[0 + 1 + 1]$
$= \frac{1}{2}(1000)(2) = 1000$

(A) Given equation: $\frac{x-a^{2}}{b+c}+\frac{x-b^{2}}{c+a}+\frac{x-c^{2}}{a+b}=4(a+b+c)$
Let $x = (a+b+c)^{2}$.
Substituting $x$ in the equation:
$\frac{(a+b+c)^{2}-a^{2}}{b+c} + \frac{(a+b+c)^{2}-b^{2}}{c+a} + \frac{(a+b+c)^{2}-c^{2}}{a+b}$
Using the identity $A^{2}-B^{2} = (A-B)(A+B)$:
$= \frac{(a+b+c-a)(a+b+c+a)}{b+c} + \frac{(a+b+c-b)(a+b+c+b)}{c+a} + \frac{(a+b+c-c)(a+b+c+c)}{a+b}$
$= \frac{(b+c)(2a+b+c)}{b+c} + \frac{(a+c)(a+2b+c)}{c+a} + \frac{(a+b)(a+b+2c)}{a+b}$
$= (2a+b+c) + (a+2b+c) + (a+b+2c)$
$= 4a+4b+4c = 4(a+b+c)$
Since the expression satisfies the equation,$x = (a+b+c)^{2}$.

(D) Given $(x-a)(x-b)=1$,we can write $(x-b) = \frac{1}{x-a}$.
We are given $a-b+5=0$,which implies $a-b = -5$,or $b-a = 5$.
We need to find the value of $(x-a)^{3}-\frac{1}{(x-a)^{3}}$.
Substitute $\frac{1}{x-a} = x-b$ into the expression:
$(x-a)^{3} - (x-b)^{3}$.
Using the identity $A^3 - B^3 = (A-B)(A^2 + AB + B^2)$ or simply substituting $x-b = x-a + (a-b)$:
Let $u = x-a$. Then $x-b = u + (a-b) = u - 5$.
So,the expression becomes $u^3 - (u-5)^3$.
Expanding this: $u^3 - (u^3 - 15u^2 + 75u - 125) = 15u^2 - 75u + 125$.
Alternatively,using the given $(x-a)(x-b)=1$:
$u(u-5) = 1 \implies u^2 - 5u = 1$.
We want $u^3 - (u-5)^3 = u^3 - (u^3 - 3u^2(5) + 3u(25) - 125) = 15u^2 - 75u + 125$.
$= 15(u^2 - 5u) + 125$.
Substituting $u^2 - 5u = 1$:
$= 15(1) + 125 = 15 + 125 = 140$.

(A) Let $x = \sqrt{2 \sqrt[3]{4 \sqrt{2 \sqrt[3]{4 \ldots}}}}$.
Squaring both sides,we get $x^{2} = 2 \sqrt[3]{4 \sqrt{2 \sqrt[3]{4 \ldots}}}$.
Now,cubing both sides,we get $(x^{2})^{3} = 2^{3} \times 4 \times \sqrt{2 \sqrt[3]{4 \ldots}}$.
This simplifies to $x^{6} = 8 \times 4 \times x$.
$x^{6} = 32x$.
Since $x \neq 0$,we divide by $x$ to get $x^{5} = 32$.
$x^{5} = 2^{5}$,which implies $x = 2$.

(B) To solve the expression,we rationalize each term individually:
$1$. $\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}} = \frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3} = \frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{3} = \sqrt{2}(\sqrt{6}-\sqrt{3}) = \sqrt{12}-\sqrt{6} = 2 \sqrt{3}-\sqrt{6}$
$2$. $\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}} = \frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2} = \frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4} = \sqrt{3}(\sqrt{6}-\sqrt{2}) = \sqrt{18}-\sqrt{6} = 3 \sqrt{2}-\sqrt{6}$
$3$. $\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2} = \sqrt{6}(\sqrt{3}-\sqrt{2}) = \sqrt{18}-\sqrt{12} = 3 \sqrt{2}-2 \sqrt{3}$
Substituting these back into the original expression:
$(2 \sqrt{3}-\sqrt{6}) - (3 \sqrt{2}-\sqrt{6}) + (3 \sqrt{2}-2 \sqrt{3})$
$= 2 \sqrt{3} - \sqrt{6} - 3 \sqrt{2} + \sqrt{6} + 3 \sqrt{2} - 2 \sqrt{3}$
$= (2 \sqrt{3} - 2 \sqrt{3}) + (-\sqrt{6} + \sqrt{6}) + (-3 \sqrt{2} + 3 \sqrt{2})$
$= 0 + 0 + 0 = 0$

(A) Given the equation: $a^{2}+b^{2}-c^{2}=2(a-b-c)-3$
Rearranging the terms: $a^{2}+b^{2}-c^{2}-2a+2b+2c+3=0$
Note: There is a sign correction required in the original equation to make it solvable as a sum of squares. Assuming the equation is $a^{2}+b^{2}+c^{2}=2(a-b-c)-3$:
$a^{2}-2a+1 + b^{2}+2b+1 + c^{2}+2c+1 = 0$
$(a-1)^{2} + (b+1)^{2} + (c+1)^{2} = 0$
Since the sum of squares is zero,each term must be zero:
$a-1=0 \Rightarrow a=1$
$b+1=0 \Rightarrow b=-1$
$c+1=0 \Rightarrow c=-1$
Now,calculate $4a-3b+5c$:
$4(1) - 3(-1) + 5(-1) = 4 + 3 - 5 = 2$

(C) Given: $2x + \frac{2}{x} = 3$
Divide the entire equation by $2$:
$x + \frac{1}{x} = \frac{3}{2}$
Now,cube both sides using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$:
$(x + \frac{1}{x})^3 = (\frac{3}{2})^3$
$x^3 + \frac{1}{x^3} + 3(x)(\frac{1}{x})(x + \frac{1}{x}) = \frac{27}{8}$
Substitute $x + \frac{1}{x} = \frac{3}{2}$ into the equation:
$x^3 + \frac{1}{x^3} + 3(1)(\frac{3}{2}) = \frac{27}{8}$
$x^3 + \frac{1}{x^3} + \frac{9}{2} = \frac{27}{8}$
$x^3 + \frac{1}{x^3} = \frac{27}{8} - \frac{9}{2} = \frac{27 - 36}{8} = -\frac{9}{8}$
Finally,calculate the value of $x^3 + \frac{1}{x^3} + 2$:
$-\frac{9}{8} + 2 = \frac{-9 + 16}{8} = \frac{7}{8}$

(A) Let $x = a^{2}-b^{2}$,$y = b^{2}-c^{2}$,and $z = c^{2}-a^{2}$.
Then,$x+y+z = (a^{2}-b^{2}) + (b^{2}-c^{2}) + (c^{2}-a^{2}) = 0$.
We know the algebraic identity: If $x+y+z = 0$,then $x^{3}+y^{3}+z^{3} = 3xyz$.
Substituting the values,we get:
$(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3} = 3(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})$.
Expanding the terms using the difference of squares formula $a^{2}-b^{2} = (a-b)(a+b)$:
$= 3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$.
Thus,$(a+b)(a-b)$ is one of the factors.

(D) Given $x = \sqrt[3]{5} + 2$.
Subtracting $2$ from both sides,we get $x - 2 = \sqrt[3]{5}$.
Now,cubing both sides of the equation:
$(x - 2)^3 = (\sqrt[3]{5})^3$
Using the algebraic identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$,we expand the left side:
$x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = 5$
$x^3 - 6x^2 + 12x - 8 = 5$
To find the value of $x^3 - 6x^2 + 12x - 13$,we subtract $5$ from both sides of the equation:
$x^3 - 6x^2 + 12x - 8 - 5 = 0$
$x^3 - 6x^2 + 12x - 13 = 0$.

(D) First,rationalize the denominator of the fraction $\frac{1}{3-\sqrt{8}}$:
$\frac{1}{3-\sqrt{8}} = \frac{1 \times (3+\sqrt{8})}{(3-\sqrt{8})(3+\sqrt{8})} = \frac{3+\sqrt{8}}{3^2 - (\sqrt{8})^2} = \frac{3+\sqrt{8}}{9-8} = 3+\sqrt{8}$.
Now,substitute this back into the original expression:
$(3+\sqrt{8}) + (3+\sqrt{8}) - (6+4\sqrt{2})$.
Since $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$,we can rewrite the expression as:
$(3+2\sqrt{2}) + (3+2\sqrt{2}) - (6+4\sqrt{2})$.
$= 6 + 4\sqrt{2} - 6 - 4\sqrt{2} = 0$.

(C) Given $x^{2}+\frac{1}{x^{2}}=83$.
We know that $(x-\frac{1}{x})^{2} = x^{2}+\frac{1}{x^{2}}-2$.
Substituting the value,we get $(x-\frac{1}{x})^{2} = 83-2 = 81$.
Since $x>1$,$x-\frac{1}{x}$ must be positive,so $x-\frac{1}{x} = \sqrt{81} = 9$.
Now,using the identity $(x-\frac{1}{x})^{3} = x^{3}-\frac{1}{x^{3}}-3(x-\frac{1}{x})$.
Substituting the values,$9^{3} = x^{3}-\frac{1}{x^{3}}-3(9)$.
$729 = x^{3}-\frac{1}{x^{3}}-27$.
Therefore,$x^{3}-\frac{1}{x^{3}} = 729+27 = 756$.