Question

If $x+\left(\frac{1}{x}\right)=3$, find $x^2+\left(\frac{1}{x^2}\right)$ and $x^3+\left(\frac{1}{x^3}\right)$

Answer 1

$x+\frac{1}{x}=3$

$\Rightarrow {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2x.\frac{1}{x}$

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

$\Rightarrow x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2=3^2-2=7$

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

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

$\Rightarrow x^3+\frac{1}{x^3}={(x+\frac{1}{x})}^3-3(x+\frac{1}{x})=3^3\left(3\right)=27-9$

$=18.$

Hence, the answer is $18.$