Simplification Questions in English

(D) Given that $(a+\frac{1}{a})^{2}=3$.
Taking the square root on both sides,we get $a+\frac{1}{a}=\sqrt{3}$.
To find $a^{3}+\frac{1}{a^{3}}$,we use the algebraic identity $(x+y)^{3} = x^{3} + y^{3} + 3xy(x+y)$.
Substituting $x=a$ and $y=\frac{1}{a}$,we get $(a+\frac{1}{a})^{3} = a^{3} + \frac{1}{a^{3}} + 3(a)(\frac{1}{a})(a+\frac{1}{a})$.
This simplifies to $(a+\frac{1}{a})^{3} = a^{3} + \frac{1}{a^{3}} + 3(a+\frac{1}{a})$.
Substituting the value $a+\frac{1}{a}=\sqrt{3}$ into the equation:
$(\sqrt{3})^{3} = a^{3} + \frac{1}{a^{3}} + 3(\sqrt{3})$.
$3\sqrt{3} = a^{3} + \frac{1}{a^{3}} + 3\sqrt{3}$.
Subtracting $3\sqrt{3}$ from both sides,we get $a^{3} + \frac{1}{a^{3}} = 0$.

(B) Given $a = 7 - 4 \sqrt{3}$.
First,find $\frac{1}{a}$ by rationalizing the denominator:
$\frac{1}{a} = \frac{1}{7 - 4 \sqrt{3}} \times \frac{7 + 4 \sqrt{3}}{7 + 4 \sqrt{3}} = \frac{7 + 4 \sqrt{3}}{49 - 16(3)} = \frac{7 + 4 \sqrt{3}}{49 - 48} = 7 + 4 \sqrt{3}$.
We need to find $a^{\frac{1}{2}} + a^{-\frac{1}{2}}$,which is $\sqrt{a} + \frac{1}{\sqrt{a}}$.
Let $x = \sqrt{a} + \frac{1}{\sqrt{a}}$.
Then $x^2 = (\sqrt{a} + \frac{1}{\sqrt{a}})^2 = a + \frac{1}{a} + 2$.
Substitute the values of $a$ and $\frac{1}{a}$:
$x^2 = (7 - 4 \sqrt{3}) + (7 + 4 \sqrt{3}) + 2 = 14 + 2 = 16$.
Since $a > 0$,$\sqrt{a} + \frac{1}{\sqrt{a}}$ must be positive,so $x = \sqrt{16} = 4$.

(B) To find the approximate value,we round the numbers to the nearest integers:
$21 + 39 \times 2.9 + 8.99 \approx 21 + 39 \times 3 + 9$
Following the order of operations $(BODMAS)$:
$= 21 + (39 \times 3) + 9$
$= 21 + 117 + 9$
$= 147$
Rounding to the nearest option,the value is approximately $148$.

(B) Given the equation: $22.9889 \div ? = 23$
Since $22.9889$ is approximately equal to $23$,we can write:
$23 \div ? = 23$
Dividing both sides by $23$,we get:
$? = \frac{23}{23} = 1$
Therefore,the value in place of the question mark is $1$.

(B) Given equation: $10^{3} \times 100^{3} + 999999999 = 10^{?} + 10^{?}$
Step $1$: Simplify the left side.
$100^{3} = (10^{2})^{3} = 10^{6}$.
So,$10^{3} \times 10^{6} = 10^{3+6} = 10^{9}$.
Step $2$: Approximate the constant.
$999999999$ is approximately $1000000000$,which is $10^{9}$.
Step $3$: Substitute back into the equation.
$10^{9} + 10^{9} = 10^{?} + 10^{?}$.
Comparing both sides,we get $? = 9$.

(B) Rationalizing each term separately:
$1$. $\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}} = \frac{3 \sqrt{2}(\sqrt{6}+\sqrt{3})}{6-3} = \frac{3(\sqrt{12}+\sqrt{6})}{3} = 2 \sqrt{3}+\sqrt{6}$
$2$. $\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}} = \frac{4 \sqrt{3}(\sqrt{6}+\sqrt{2})}{6-2} = \frac{4(\sqrt{18}+\sqrt{6})}{4} = 3 \sqrt{2}+\sqrt{6}$
$3$. $\frac{6}{\sqrt{8}+\sqrt{12}} = \frac{6(\sqrt{12}-\sqrt{8})}{12-8} = \frac{6(2 \sqrt{3}-2 \sqrt{2})}{4} = \frac{3(2 \sqrt{3}-2 \sqrt{2})}{2} = 3 \sqrt{3}-3 \sqrt{2}$
Substituting these values into the original expression:
$(2 \sqrt{3}+\sqrt{6}) - (3 \sqrt{2}+\sqrt{6}) - (3 \sqrt{3}-3 \sqrt{2})$
$= 2 \sqrt{3} + \sqrt{6} - 3 \sqrt{2} - \sqrt{6} - 3 \sqrt{3} + 3 \sqrt{2}$
$= (2 \sqrt{3} - 3 \sqrt{3}) + (\sqrt{6} - \sqrt{6}) + (-3 \sqrt{2} + 3 \sqrt{2})$
$= -\sqrt{3}$

(D) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.
First,we find $x^{2}+y^{2}+z^{2}$ using the identity $(x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$.
Substituting the given values: $(1)^{2} = x^{2}+y^{2}+z^{2} + 2(-1) \Rightarrow 1 = x^{2}+y^{2}+z^{2} - 2 \Rightarrow x^{2}+y^{2}+z^{2} = 3$.
Now,substitute all values into the main identity:
$x^{3}+y^{3}+z^{3} - 3(-1) = (1)(3 - (-1))$.
$x^{3}+y^{3}+z^{3} + 3 = 1(3+1) = 4$.
$x^{3}+y^{3}+z^{3} = 4 - 3 = 1$.

(C) Given the equation: $x^{2}+y^{2}+z^{2}=xy+yz+zx$
Multiply both sides by $2$: $2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx=0$
This can be rewritten as: $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$
Since the sum of squares is zero,each term must be zero:
$x-y=0 \Rightarrow x=y$
$y-z=0 \Rightarrow y=z$
$z-x=0 \Rightarrow z=x$
Thus,$x=y=z$.
Substitute $y=x$ and $z=x$ into the expression $\frac{4x+2y-3z}{2x}$:
$\frac{4x+2(x)-3(x)}{2x} = \frac{4x+2x-3x}{2x} = \frac{3x}{2x} = \frac{3}{2}$.

(C) Given equation: $x(3 - \frac{2}{x}) = \frac{3}{x}$
Expanding the left side: $3x - 2 = \frac{3}{x}$
Rearranging the terms: $3x - \frac{3}{x} = 2$
Dividing both sides by $3$: $x - \frac{1}{x} = \frac{2}{3}$
Squaring both sides: $(x - \frac{1}{x})^2 = (\frac{2}{3})^2$
$x^2 + \frac{1}{x^2} - 2 = \frac{4}{9}$
Adding $2$ to both sides: $x^2 + \frac{1}{x^2} = \frac{4}{9} + 2$
$x^2 + \frac{1}{x^2} = \frac{4 + 18}{9} = \frac{22}{9}$
Converting to mixed fraction: $\frac{22}{9} = 2 \frac{4}{9}$

(A) Given equation: $x^{2}+y^{2}+z^{2}+2=2(y-x)$
Rearranging the terms: $x^{2}+2x+y^{2}-2y+z^{2}+2=0$
Completing the squares: $(x^{2}+2x+1)+(y^{2}-2y+1)+z^{2}=0$
This simplifies to: $(x+1)^{2}+(y-1)^{2}+z^{2}=0$
Since the sum of squares is zero,each term must be zero:
$x+1=0 \Rightarrow x=-1$
$y-1=0 \Rightarrow y=1$
$z=0$
Therefore,$x^{3}+y^{3}+z^{3} = (-1)^{3} + (1)^{3} + (0)^{3} = -1 + 1 + 0 = 0$.

(B) Given the equations $a^{3} b = 180$ and $a b c = 180$.
Since $a, b, c$ are positive integers,we equate the two expressions: $a^{3} b = a b c$.
Dividing both sides by $ab$ (since $a, b > 0$),we get $a^{2} = c$.
Now,we factorize $180$: $180 = 2^{2} \times 3^{2} \times 5 = 4 \times 9 \times 5$.
We need to find $a$ such that $a^{2}$ is a factor of $180$ and $a^{3} b = 180$.
If $a = 1$,then $c = a^{2} = 1^{2} = 1$. Then $1^{3} \times b = 180$,so $b = 180$. This satisfies the condition.
If $a = 2$,then $c = a^{2} = 2^{2} = 4$. Then $2^{3} \times b = 180$,which gives $8b = 180$,so $b = 22.5$ (not an integer).
If $a = 3$,then $c = a^{2} = 3^{2} = 9$. Then $3^{3} \times b = 180$,which gives $27b = 180$,so $b = 180/27 = 20/3$ (not an integer).
Thus,the only integer solution is $a = 1, b = 180, c = 1$.

(A) Given $(x+\frac{1}{x})^{2}=3$.
Taking the square root on both sides,we get $x+\frac{1}{x}=\sqrt{3}$.
Cubing both sides,we get $(x+\frac{1}{x})^{3}=(\sqrt{3})^{3}$.
Using the identity $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$,we have $x^{3}+\frac{1}{x^{3}}+3(x+\frac{1}{x})=3\sqrt{3}$.
Substituting $x+\frac{1}{x}=\sqrt{3}$,we get $x^{3}+\frac{1}{x^{3}}+3\sqrt{3}=3\sqrt{3}$,which implies $x^{3}+\frac{1}{x^{3}}=0$.
Multiplying by $x^{3}$,we get $x^{6}+1=0$,so $x^{6}=-1$.
Now,substitute $x^{6}=-1$ into the expression $(x^{72}+x^{66}+x^{54}+x^{36}+x^{24}+x^{6}+1)$:
$= (x^{6})^{12} + (x^{6})^{11} + (x^{6})^{9} + (x^{6})^{6} + (x^{6})^{4} + x^{6} + 1$
$= (-1)^{12} + (-1)^{11} + (-1)^{9} + (-1)^{6} + (-1)^{4} + (-1) + 1$
$= 1 - 1 - 1 + 1 + 1 - 1 + 1 = 1$.

(C) Given that $a+b+c=0.$
$\Rightarrow b+c=-a$
Squaring both sides:
$\Rightarrow (b+c)^{2}=(-a)^{2}$
$\Rightarrow b^{2}+c^{2}+2bc=a^{2}$
Now,add $a^{2}$ to both sides to match the numerator:
$\Rightarrow a^{2}+b^{2}+c^{2}+2bc=2a^{2}$
$\Rightarrow a^{2}+b^{2}+c^{2}=2a^{2}-2bc$
$\Rightarrow a^{2}+b^{2}+c^{2}=2(a^{2}-bc)$
Dividing both sides by $(a^{2}-bc)$:
$\Rightarrow \frac{a^{2}+b^{2}+c^{2}}{a^{2}-bc}=2$

(B) Given $n = 7 + 4 \sqrt{3}$.
We can rewrite $n$ as $n = 4 + 3 + 2 \times 2 \times \sqrt{3} = 2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})$.
Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,we get $n = (2 + \sqrt{3})^2$.
Therefore,$\sqrt{n} = 2 + \sqrt{3}$.
Now,find $\frac{1}{\sqrt{n}} = \frac{1}{2 + \sqrt{3}}$.
Rationalizing the denominator: $\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$.
Finally,$\sqrt{n} + \frac{1}{\sqrt{n}} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.

(B) We know the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$.
Substituting the given values: $6^{2} = 14 + 2(ab+bc+ca)$.
$36 = 14 + 2(ab+bc+ca) \Rightarrow 2(ab+bc+ca) = 22 \Rightarrow ab+bc+ca = 11$.
Next,we use the identity $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-(ab+bc+ca))$.
Substituting the known values: $36 - 3abc = 6(14 - 11)$.
$36 - 3abc = 6(3) = 18$.
$3abc = 36 - 18 = 18$.
$abc = 6$.

(D) Given the equation: $(a-1) \sqrt{2} + 3 = b \sqrt{2} + a$.
Since $a$ and $b$ are rational numbers,we can equate the rational and irrational parts on both sides of the equation.
The rational part on the left is $3$ and on the right is $a$. Therefore,$a = 3$.
The coefficient of the irrational part $\sqrt{2}$ on the left is $(a-1)$ and on the right is $b$. Therefore,$a - 1 = b$.
Substituting $a = 3$ into the second equation: $3 - 1 = b$,which gives $b = 2$.
Finally,the value of $(a+b) = 3 + 2 = 5$.

(A) Given $\left(x+\frac{1}{x}\right)^{2}=3$.
Taking the square root on both sides,we get $x+\frac{1}{x}=\sqrt{3}$.
Cubing both sides,we get $\left(x+\frac{1}{x}\right)^{3} = (\sqrt{3})^{3}$.
Using the identity $(a+b)^{3} = a^{3}+b^{3}+3ab(a+b)$,we have $x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right) = 3\sqrt{3}$.
Substituting $x+\frac{1}{x}=\sqrt{3}$,we get $x^{3}+\frac{1}{x^{3}}+3(\sqrt{3}) = 3\sqrt{3}$.
This simplifies to $x^{3}+\frac{1}{x^{3}} = 0$,which implies $x^{6}+1 = 0$.
Now,the given expression is $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1$.
Grouping the terms,we get $x^{200}(x^{6}+1)+x^{84}(x^{6}+1)+x^{12}(x^{6}+1)+1(x^{6}+1)$.
Since $x^{6}+1=0$,the entire expression becomes $x^{200}(0)+x^{84}(0)+x^{12}(0)+1(0) = 0$.

(D) Given equation: $4003 \times 77 - 21015 = ? \times 116$
Step $1$: Calculate the product $4003 \times 77$.
$4003 \times 77 = 308231$
Step $2$: Subtract $21015$ from the result.
$308231 - 21015 = 287216$
Step $3$: Solve for $?$.
$? \times 116 = 287216$
$? = \frac{287216}{116}$
$? = 2476$

(B) To find the value of the expression, we perform the division for each term:
$1. (4444 \div 40) = 111.1$
$2. (645 \div 25) = 25.8$
$3. (3991 \div 26) = 153.5$
Now, add these results together:
$? = 111.1 + 25.8 + 153.5$
$? = 290.4$

(B) First,convert the mixed fractions into improper fractions:
$5 \frac{17}{37} = \frac{5 \times 37 + 17}{37} = \frac{185 + 17}{37} = \frac{202}{37}$
$4 \frac{51}{52} = \frac{4 \times 52 + 51}{52} = \frac{208 + 51}{52} = \frac{259}{52}$
$11 \frac{1}{7} = \frac{11 \times 7 + 1}{7} = \frac{77 + 1}{7} = \frac{78}{7}$
$2 \frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4}$
Now,substitute these into the expression:
$? = \frac{202}{37} \times \frac{259}{52} \times \frac{78}{7} + \frac{11}{4}$
Simplify the multiplication:
$= \frac{202}{37} \times \frac{259}{7} \times \frac{78}{52} + \frac{11}{4}$
$= \frac{202}{37} \times 37 \times \frac{3}{2} + \frac{11}{4}$
$= 202 \times \frac{3}{2} + \frac{11}{4}$
$= 101 \times 3 + 2.75$
$= 303 + 2.75 = 305.75$

(C) To solve the expression $\frac{5}{8} \times 4011.33 + \frac{7}{10} \times 3411.22$,we perform the following steps:
Step $1$: Calculate $\frac{5}{8} \times 4011.33 = 0.625 \times 4011.33 = 2507.08125$.
Step $2$: Calculate $\frac{7}{10} \times 3411.22 = 0.7 \times 3411.22 = 2387.854$.
Step $3$: Add the two results: $2507.08125 + 2387.854 = 4894.93525$.
Rounding to the nearest option,we get $4895 \approx 4890$.

(A) Given expression: $335.01 \times 244.99 \div 55 = ?$
Rounding the values to the nearest integers,we get:
$335 \times 245 \div 55 = ?$
Using the order of operations $(BODMAS)$:
$= 335 \times (245 \div 55)$
$= 335 \times 4.4545...$
$= 1492.27...$
Rounding the result to the nearest option,we get $1490$.

(B) To solve the expression $\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$,we rationalize the denominator for each term.
First term: $\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})} = \frac{3 \sqrt{6}-3 \sqrt{12}}{3-6} = \frac{3 \sqrt{6}-6 \sqrt{3}}{-3} = -\sqrt{6}+2 \sqrt{3}$.
Second term: $\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} = \frac{4 \sqrt{18}-4 \sqrt{6}}{6-2} = \frac{12 \sqrt{2}-4 \sqrt{6}}{4} = 3 \sqrt{2}-\sqrt{6}$.
Third term: $\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} = \frac{\sqrt{18}-\sqrt{12}}{3-2} = 3 \sqrt{2}-2 \sqrt{3}$.
Combining these: $(-\sqrt{6}+2 \sqrt{3}) - (3 \sqrt{2}-\sqrt{6}) + (3 \sqrt{2}-2 \sqrt{3})$.
$= -\sqrt{6}+2 \sqrt{3}-3 \sqrt{2}+\sqrt{6}+3 \sqrt{2}-2 \sqrt{3} = 0$.

(A) First,simplify the numerator: $2 \frac{1}{3} - 1 \frac{2}{11} = \frac{7}{3} - \frac{13}{11} = \frac{77 - 39}{33} = \frac{38}{33}$.
Next,simplify the denominator: $3 + \frac{1}{3 + \frac{1}{3 + \frac{1}{3}}} = 3 + \frac{1}{3 + \frac{1}{\frac{10}{3}}} = 3 + \frac{1}{3 + \frac{3}{10}} = 3 + \frac{1}{\frac{33}{10}} = 3 + \frac{10}{33} = \frac{99 + 10}{33} = \frac{109}{33}$.
Finally,divide the numerator by the denominator: $\frac{38/33}{109/33} = \frac{38}{33} \times \frac{33}{109} = \frac{38}{109}$.

(B) Given expression: $3+\frac{1}{\sqrt{3}}+\frac{1}{3+\sqrt{3}}+\frac{1}{\sqrt{3}-3}$
We can rewrite the last term as: $\frac{1}{\sqrt{3}-3} = -\frac{1}{3-\sqrt{3}}$
So the expression becomes: $3+\frac{1}{\sqrt{3}}+\left(\frac{1}{3+\sqrt{3}}-\frac{1}{3-\sqrt{3}}\right)$
Now,simplify the term in the parentheses by taking the common denominator $(3+\sqrt{3})(3-\sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9-3 = 6$:
$\frac{(3-\sqrt{3}) - (3+\sqrt{3})}{6} = \frac{3-\sqrt{3}-3-\sqrt{3}}{6} = \frac{-2\sqrt{3}}{6} = -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}$
Substituting this back into the original expression:
$3 + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}} = 3$

(A) Given the equation: $x+\frac{2}{3+\frac{4}{5+\frac{7}{6}}}=10$
First,simplify the innermost fraction: $5+\frac{7}{6} = \frac{30+7}{6} = \frac{37}{6}$
Substitute this back into the expression: $x+\frac{2}{3+\frac{4}{\frac{37}{6}}} = 10$
Simplify the complex fraction: $x+\frac{2}{3+\frac{4 \times 6}{37}} = 10$
$x+\frac{2}{3+\frac{24}{37}} = 10$
Simplify the denominator: $3+\frac{24}{37} = \frac{111+24}{37} = \frac{135}{37}$
Substitute back: $x+\frac{2}{\frac{135}{37}} = 10$
$x+\frac{2 \times 37}{135} = 10$
$x+\frac{74}{135} = 10$
Solve for $x$: $x = 10 - \frac{74}{135}$
$x = \frac{1350 - 74}{135} = \frac{1276}{135}$

(B) To solve the expression $3+\frac{3}{3+\frac{1}{3+\frac{1}{3}}}$,we start from the bottom of the fraction.
Step $1$: Simplify the lowest part,$3+\frac{1}{3} = \frac{9+1}{3} = \frac{10}{3}$.
Step $2$: Substitute this back into the expression: $3+\frac{3}{3+\frac{1}{10/3}} = 3+\frac{3}{3+\frac{3}{10}}$.
Step $3$: Simplify the denominator $3+\frac{3}{10} = \frac{30+3}{10} = \frac{33}{10}$.
Step $4$: Substitute this back: $3+\frac{3}{33/10} = 3+3 \times \frac{10}{33} = 3+\frac{10}{11}$.
Step $5$: Final calculation: $3+\frac{10}{11} = \frac{33+10}{11} = \frac{43}{11}$.

(C) Given $x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}}$.
Rationalizing the denominator inside the square root:
$x=\sqrt{\frac{(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}} = \sqrt{\frac{(\sqrt{5}+1)^{2}}{5-1}} = \sqrt{\frac{(\sqrt{5}+1)^{2}}{4}} = \frac{\sqrt{5}+1}{2}$.
Now,substitute $x = \frac{\sqrt{5}+1}{2}$ into the expression $5x^{2}-5x-1$:
$5\left(\frac{\sqrt{5}+1}{2}\right)^{2} - 5\left(\frac{\sqrt{5}+1}{2}\right) - 1$
$= 5\left(\frac{5+1+2\sqrt{5}}{4}\right) - \frac{5\sqrt{5}+5}{2} - 1$
$= 5\left(\frac{6+2\sqrt{5}}{4}\right) - \frac{5\sqrt{5}+5}{2} - 1$
$= 5\left(\frac{3+\sqrt{5}}{2}\right) - \frac{5\sqrt{5}+5}{2} - 1$
$= \frac{15+5\sqrt{5}-5\sqrt{5}-5-2}{2} = \frac{8}{2} = 4$.

(C) To solve the expression $67.99 \%$ of $1401 - 13.99 \%$ of $1299$,we can approximate the values for easier calculation.
$67.99 \% \approx 68 \%$
$1401 \approx 1400$
$13.99 \% \approx 14 \%$
$1299 \approx 1300$
Now,substitute these values into the expression:
$? = (68 \% \text{ of } 1400) - (14 \% \text{ of } 1300)$
Calculate each part:
$68 \% \text{ of } 1400 = \frac{68}{100} \times 1400 = 68 \times 14 = 952$
$14 \% \text{ of } 1300 = \frac{14}{100} \times 1300 = 14 \times 13 = 182$
Subtract the results:
$? = 952 - 182 = 770$
Thus,the correct option is $C$.

(A) Given expression: $? = \left(\frac{24}{9}\right)^{2} \times \frac{399}{39} \div \frac{41}{899}$
Step $1$: Simplify the fraction $\frac{24}{9}$ to $\frac{8}{3}$.
Step $2$: Calculate the square: $\left(\frac{8}{3}\right)^{2} = \frac{64}{9}$.
Step $3$: Simplify $\frac{399}{39} \approx 10.23$ and $\frac{41}{899} \approx 0.0456$.
Step $4$: Convert division to multiplication: $? = \frac{64}{9} \times \frac{399}{39} \times \frac{899}{41}$.
Step $5$: Calculate the values: $\frac{64}{9} \approx 7.111$,$\frac{399}{39} \approx 10.2307$,$\frac{899}{41} \approx 21.9268$.
Step $6$: Multiply these values: $7.111 \times 10.2307 \times 21.9268 \approx 1594.36$.
Rounding to the nearest option,the value is approximately $1600$.

(B) Given equation: $(15 \times 0.40)^{4} \div (1080 \div 30)^{4} \times (27 \times 8)^{4} = (3 \times 2)^{?+5}$
Step $1$: Simplify the terms inside the parentheses.
$(15 \times 0.40) = 6$
$(1080 \div 30) = 36$
$(27 \times 8) = 216$
Step $2$: Substitute these values into the equation.
$(6)^{4} \div (36)^{4} \times (216)^{4} = (6)^{?+5}$
Step $3$: Express all bases as powers of $6$.
$36 = 6^{2}$ and $216 = 6^{3}$
So,$(6)^{4} \div (6^{2})^{4} \times (6^{3})^{4} = (6)^{?+5}$
$(6)^{4} \div (6)^{8} \times (6)^{12} = (6)^{?+5}$
Step $4$: Use exponent laws $a^{m} \div a^{n} = a^{m-n}$ and $a^{m} \times a^{n} = a^{m+n}$.
$(6)^{4-8+12} = (6)^{?+5}$
$(6)^{8} = (6)^{?+5}$
Step $5$: Equate the exponents.
$8 = ? + 5$
$? = 8 - 5 = 3$

(D) Let the missing value be $x$. The equation is $\frac{x^2}{10} + 1 \frac{5}{12} = 3 \frac{1}{4} + 2 \frac{1}{2} - 1 \frac{5}{6}$.
Convert mixed fractions to improper fractions:
$\frac{x^2}{10} + \frac{17}{12} = \frac{13}{4} + \frac{5}{2} - \frac{11}{6}$.
Find a common denominator $(12)$ for the right side:
$\frac{x^2}{10} + \frac{17}{12} = \frac{39}{12} + \frac{30}{12} - \frac{22}{12} = \frac{47}{12}$.
Isolate the term with $x$:
$\frac{x^2}{10} = \frac{47}{12} - \frac{17}{12} = \frac{30}{12} = \frac{5}{2}$.
Solve for $x^2$:
$x^2 = \frac{5}{2} \times 10 = 25$.
Therefore,$x = \sqrt{25} = 5$.

(C) Given equation: $(?)^{3} + \sqrt{49} = 92 \times 576 \div 2 \sqrt{1296}$
Step $1$: Simplify the square roots.
$\sqrt{1296} = 36$ and $\sqrt{49} = 7$.
Step $2$: Substitute these values into the equation.
$(?)^{3} + 7 = 92 \times 576 \div (2 \times 36)$
Step $3$: Perform the division.
$(?)^{3} + 7 = 92 \times 576 \div 72$
$(?)^{3} + 7 = 92 \times 8$
Step $4$: Multiply the values.
$(?)^{3} + 7 = 736$
Step $5$: Solve for $(?)^{3}$.
$(?)^{3} = 736 - 7 = 729$
Step $6$: Find the cube root.
$? = \sqrt[3]{729} = 9$
Therefore,the correct option is $C$.

(C) Let the missing value be $x$.
The given equation is: $85 + x = \frac{1}{6} \times \frac{92}{100} \times \frac{24}{23} \times 650$
Step $1$: Simplify the fraction $1 \frac{1}{23}$ as $\frac{24}{23}$.
Step $2$: Calculate the right-hand side:
$\frac{1}{6} \times \frac{92}{100} \times \frac{24}{23} \times 650$
$= (\frac{1}{6} \times 24) \times (\frac{92}{23}) \times (\frac{650}{100})$
$= 4 \times 4 \times 6.5$
$= 16 \times 6.5 = 104$
Step $3$: Solve for $x$:
$85 + x = 104$
$x = 104 - 85$
$x = 19$

(C) Price of one pen $= Rs. 7$
Price of one packet of wax colours $= Rs. 22$
Price of one calculator $= Rs. 175$
Price of one pencil box $= (7 + 22) + 14 = 29 + 14 = Rs. 43$
Total amount paid $= (20 \times 7) + (8 \times 22) + (6 \times 175) + (7 \times 43)$
Total amount paid $= 140 + 176 + 1050 + 301 = Rs. 1667$

(C) Given expression $= \frac{(81)^{3.6} \times (9)^{2.7}}{(81)^{4.2} \times 3}$
Expressing all terms with base $3$:
$= \frac{(3^4)^{3.6} \times (3^2)^{2.7}}{(3^4)^{4.2} \times 3^1}$
Using the exponent rule $(a^m)^n = a^{m \times n}$:
$= \frac{3^{14.4} \times 3^{5.4}}{3^{16.8} \times 3^1}$
Using the rule $a^m \times a^n = a^{m+n}$:
$= \frac{3^{14.4 + 5.4}}{3^{16.8 + 1}} = \frac{3^{19.8}}{3^{17.8}}$
Using the rule $a^m \div a^n = a^{m-n}$:
$= 3^{19.8 - 17.8} = 3^2 = 9$

(C) Let the marked price of the article be $Rs. x$ and the cost price $(CP)$ be $Rs. 100$.
According to the problem,the selling price $(SP)$ after a $40 \%$ discount is:
$SP = x - 0.40x = 0.60x$
Given that there is a loss of $30 \%$,the $SP$ is also equal to $70 \%$ of the $CP$:
$SP = 100 - 30 = 70$
Equating the two expressions for $SP$:
$0.60x = 70$
$x = \frac{70}{0.60} = \frac{700}{6} = \frac{350}{3}$
If the article is sold at the marked price $(x)$,the profit is:
$Profit = SP - CP = \frac{350}{3} - 100 = \frac{350 - 300}{3} = \frac{50}{3} = 16 \frac{2}{3}$
Since the $CP$ is $100$,the profit percentage is $16 \frac{2}{3} \%$.

(C) Given equation: $a^{2}+b^{2}+c^{2}=2(a-b-c)-3$
Expand the right side: $a^{2}+b^{2}+c^{2}=2a-2b-2c-3$
Rearrange the terms to one side: $a^{2}-2a+b^{2}+2b+c^{2}+2c+3=0$
Split $3$ into $1+1+1$: $(a^{2}-2a+1)+(b^{2}+2b+1)+(c^{2}+2c+1)=0$
Rewrite as perfect squares: $(a-1)^{2}+(b+1)^{2}+(c+1)^{2}=0$
Since the sum of squares is zero,each term must be zero: $a-1=0, b+1=0, c+1=0$
Solving these gives: $a=1, b=-1, c=-1$
Calculate $(a-b+c)$: $1-(-1)+(-1) = 1+1-1 = 1$

(A) Given equation: $x^{2}+3x+1=0$
Dividing the entire equation by $x$ (assuming $x \neq 0$):
$x+3+\frac{1}{x}=0$
Rearranging the terms,we get:
$x+\frac{1}{x}=-3$
Now,cubing both sides of the equation:
$\left(x+\frac{1}{x}\right)^{3}=(-3)^{3}$
Using the algebraic identity $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$:
$x^{3}+\frac{1}{x^{3}}+3\left(x\cdot\frac{1}{x}\right)\left(x+\frac{1}{x}\right)=-27$
Substituting the value of $(x+\frac{1}{x})=-3$:
$x^{3}+\frac{1}{x^{3}}+3(1)(-3)=-27$
$x^{3}+\frac{1}{x^{3}}-9=-27$
$x^{3}+\frac{1}{x^{3}}=-27+9$
$x^{3}+\frac{1}{x^{3}}=-18$

(D) Given the equation $x^{a} \cdot x^{b} \cdot x^{c} = 1$.
Using the laws of exponents,$x^{a+b+c} = 1$.
Since $1 = x^{0}$,we have $x^{a+b+c} = x^{0}$.
Equating the exponents,we get $a + b + c = 0$.
We know the algebraic identity: $a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
Substituting $a + b + c = 0$ into the identity:
$a^{3} + b^{3} + c^{3} - 3abc = (0)(a^{2} + b^{2} + c^{2} - ab - bc - ca) = 0$.
Therefore,$a^{3} + b^{3} + c^{3} = 3abc$.

(B) Given the equation: $a+\frac{1}{a}+2=0$
Multiply the entire equation by $a$ to clear the denominator:
$a^2 + 1 + 2a = 0$
This can be rewritten as:
$a^2 + 2a + 1 = 0$
$(a+1)^2 = 0$
Taking the square root on both sides:
$a+1 = 0 \Rightarrow a = -1$
Now,substitute $a = -1$ into the expression $\left(a^{37}-\frac{1}{a^{100}}\right)$:
$= (-1)^{37} - \frac{1}{(-1)^{100}}$
Since $-1$ raised to an odd power is $-1$ and $-1$ raised to an even power is $1$:
$= -1 - \frac{1}{1}$
$= -1 - 1 = -2$

(B) Given that $a+b+c=0$.
This implies $(a+c) = -b$.
Squaring both sides,we get $(a+c)^{2} = (-b)^{2}$.
$a^{2} + c^{2} + 2ac = b^{2}$.
Therefore,$a^{2} + c^{2} = b^{2} - 2ac$.
Now,substitute this into the expression $\frac{a^{2}+b^{2}+c^{2}}{b^{2}-ca}$:
$\frac{(a^{2}+c^{2}) + b^{2}}{b^{2}-ca} = \frac{(b^{2}-2ac) + b^{2}}{b^{2}-ca}$.
$= \frac{2b^{2}-2ac}{b^{2}-ca} = \frac{2(b^{2}-ac)}{b^{2}-ac}$.
Since $b^{2} \neq ca$,we can cancel the term $(b^{2}-ac)$,which gives the result $2$.

(D) We know the algebraic identity: $a^{4}+a^{2} b^{2}+b^{4} = (a^{2}+a b+b^{2})(a^{2}-a b+b^{2})$.
Given $a^{4}+a^{2} b^{2}+b^{4} = 8$ and $a^{2}+a b+b^{2} = 4$.
Substituting these values into the identity:
$8 = 4(a^{2}-a b+b^{2})$
$a^{2}-a b+b^{2} = 2$ ........$(1)$
We are also given $a^{2}+a b+b^{2} = 4$ ........$(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(a^{2}+a b+b^{2}) - (a^{2}-a b+b^{2}) = 4 - 2$
$a^{2}+a b+b^{2}-a^{2}+a b-b^{2} = 2$
$2ab = 2$
$ab = 1$.

(D) We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
This can also be written as: $a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$.
Substituting this into the given expression:
$\frac{\frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}} = \frac{1}{2}(a+b+c)$.
Given $a=25, b=15, c=-10$,we have:
$\frac{1}{2}(25+15-10) = \frac{1}{2}(30) = 15$.

(A) Given equation: $3463 \times 295 - 18611 = ? + 5883$
Step $1$: Calculate the product $3463 \times 295$.
$3463 \times 295 = 1021585$
Step $2$: Substitute the value back into the equation.
$1021585 - 18611 = ? + 5883$
Step $3$: Subtract $18611$ from $1021585$.
$1021585 - 18611 = 1002974$
Step $4$: Solve for $?$.
$? = 1002974 - 5883$
$? = 997091$

(C) First,calculate the squares of the given numbers:
$(23.1)^{2} = 533.61$
$(48.6)^{2} = 2361.96$
$(39.8)^{2} = 1584.04$
Substitute these values into the equation:
$533.61 + 2361.96 - 1584.04 = ? + 1147.69$
Simplify the left side:
$2895.57 - 1584.04 = ? + 1147.69$
$1311.53 = ? + 1147.69$
Solve for $?$:
$? = 1311.53 - 1147.69$
$? = 163.84$

(D) To solve the expression $\frac{28}{65} \times \frac{195}{308} \div \frac{39}{44} + \frac{5}{26}$,we follow the $BODMAS$ rule.
First,convert the division into multiplication by taking the reciprocal of the fraction: $\frac{28}{65} \times \frac{195}{308} \times \frac{44}{39} + \frac{5}{26}$.
Simplify the terms: $\frac{28}{308} = \frac{1}{11}$,$\frac{195}{65} = 3$,and $\frac{44}{39}$ remains.
Calculating the product: $(\frac{1}{11} \times 3 \times \frac{44}{39}) = (3 \times \frac{4}{39}) = \frac{4}{13}$.
Now,add the remaining fraction: $\frac{4}{13} + \frac{5}{26} = \frac{8}{26} + \frac{5}{26} = \frac{13}{26} = \frac{1}{2}$.

(A) To find the approximate value,we round off the numbers to their nearest integers:
$43931.03 \approx 43931$
$2111.02 \approx 2111$
$401.04 \approx 401$
Now,the expression becomes:
$43931 \div 2111 \times 401 = ?$
First,perform the division:
$43931 \div 2111 \approx 20.81$
Then,multiply by $401$:
$20.81 \times 401 \approx 8344.81$
Alternatively,using estimation for quick calculation:
$44000 \div 2000 \times 400 = 22 \times 400 = 8800$
Comparing the result with the given options,the closest value is $8800$.

(C) To solve the expression $59.88 \div 12.21 \times 6.35$,we can approximate the values to the nearest whole numbers for a quick estimation:
$59.88 \approx 60$
$12.21 \approx 12$
$6.35 \approx 6$
Now,substitute these values into the expression:
$60 \div 12 \times 6$
Following the order of operations ($BODMAS$/$PEMDAS$),perform division first:
$60 \div 12 = 5$
Then,perform multiplication:
$5 \times 6 = 30$
Thus,the value is approximately $30$.

(B) To find the value,we perform the multiplication as follows:
$\frac{1}{8} \times \frac{2}{3} \times \frac{3}{5} \times 1715$
First,simplify the fractions:
$= \frac{1 \times 2 \times 3}{8 \times 3 \times 5} \times 1715$
$= \frac{6}{120} \times 1715$
$= \frac{1}{20} \times 1715$
$= \frac{1715}{20} = 85.75$
Rounding to the nearest integer,we get $85$.