Question

If $\left(x+\frac{1}{x}\right)=4$, find the value of

(1) $\left(x^3+\frac{1}{x^3}\right)$

(2) $\left(x-\frac{1}{x}\right)$

(3) $\left(x^3-\frac{1}{x^3}\right)$

Answer 1

Given:

$\left(x+\frac{1}{x}\right)=4$

$\Rightarrow {\left(x+\frac{1}{x}\right)}^2=4^2$ [$\because (a+b{)}^2=a^2+b^2+2ab$]

$\Rightarrow x^2+\frac{1}{x^2}+2x^2\frac{1}{x^2}=16$

$\Rightarrow x^2+\frac{1}{x^2}=14$

(1)

$\left(x^3+\frac{1}{x^3}\right)=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+x^2\times \frac{1}{x^2}\right)$

$=\left(4\right)\left(14+1\right)$

$=60$

$\therefore \left(x^3+\frac{1}{x^3}\right)=60$.

(2)

$\left(x+\frac{1}{x}\right)=4$

$\Rightarrow {\left(x-\frac{1}{x}\right)}^2={\left(x+\frac{1}{x}\right)}^2-4$

$\Rightarrow {\left(x-\frac{1}{x}\right)}^2=4^2-4$

$\Rightarrow {\left(x-\frac{1}{x}\right)}^2=16-4$ [$\because x+\frac{1}{x}=4$]

$\Rightarrow {\left(x-\frac{1}{x}\right)}^2=12$

$\Rightarrow \left(x-\frac{1}{x}\right)=\pm 2\sqrt{3}$

(3)

$\left(x^3-\frac{1}{x^3}\right)=\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+1\right)$

$=2\sqrt{3}(2-1)$ [$\because x-\frac{1}{x}=2\sqrt{3}$]

$=2\sqrt{3}$