Question

if x+$\frac{1}{x}$=$\sqrt{11}$, find the value of $x^{3}$+$\frac{1}{x^{3}}$

Answer 1

Answer:

$8 \sqrt{11}$

Answer 2

Answer:

$8\sqrt{11}$

Step-by-step explanation:

Given:

$x+\dfrac{1}{x}=\sqrt{11}$

Sum of two squares

$\boxed{a^2+b^2=(a+b)^2-2ab}$

$\textsf{Let}\; a = x$

$\textsf{Let}\; b = \dfrac{1}{x}$

Therefore:

$\begin{aligned}x^2+\dfrac{1}{x^2}&=\left(x+\dfrac{1}{x}\right)^2-2x\left(\dfrac{1}{x}\right)\\\\&=\left(\sqrt{11}\right)^2-2\\\\&=11-2\\\\&=9\end{aligned}$

Sum of two cubes

$\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}$

$\textsf{Let}\; a = x$

$\textsf{Let}\; b = \dfrac{1}{x}$

Therefore:

$\begin{aligned}x^3+\dfrac{1}{x^3}&=\left(x+\dfrac{1}{x}\right)\left(x^2-x\left(\dfrac{1}{x}\right)+\left(\dfrac{1}{x}\right)^2\right)\\\\ &=\left(x+\dfrac{1}{x}\right)\left(x^2-1+\dfrac{1}{x^2}\right)\\\\&=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right)\\\\&=\sqrt{11}\left(9-1\right)\\\\&=8\sqrt{11}\end{aligned}$