Question

Choose the correct answer from the alternatives given.
If x - $\frac{1}{x}$ = 3 then find the value of $x^3+\frac{1}{x^3}$.

• A. 10$\sqrt{13}$
• B. 100$\sqrt{3}$
• C. 13$\sqrt{10}$
• D. 130$\sqrt{10}$

Answer 1

The correct option is A 10$\sqrt{13}$

We should determine the value of x+ $\frac{1}{x}$- first.
We know that,
${(x+\frac{1}{x})}^2={(x-\frac{1}{x})}^2+4(x\times \frac{1}{x})$
= $(3{)}^2$ + 4 = 9 + 4 = 13
$x+\frac{1}{x}=\pm \sqrt{13}$
cubing both sides
${(x+\frac{1}{x})}^3=(\sqrt{13}{)}^3$
$x^3+\frac{1}{x^3}+3\times x\times \frac{1}{x}(x+\frac{1}{x})=13\sqrt{13}$
$x^3+\frac{1}{x^3}+3\sqrt{13}=13\sqrt{13}$

$\Rightarrow x^3+\frac{1}{x^3}=13\sqrt{13}-3\sqrt{13}$
$x^3+\frac{1}{x^3}=10\sqrt{13}$