Question

If $x+\frac{1}{x}=7$ then find the value of $x^3+\frac{1}{x^3}$

Answer 1

Given: $x+\frac{1}{x}=7\dots \dots \left(i\right)$

on Cubing both the sides, we get,

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

$\Rightarrow x^3+\frac{1}{x^3}+3\left(x\right)\left(\frac{1}{x}\right)(x+\frac{1}{x})=343$ [$\because (a+b{)}^3=a^3+b^3+3ab(a+b)$]

$\Rightarrow x^3+\frac{1}{x^3}+3\left(7\right)=343$ [Using $eq.\left(i\right)$]

$\Rightarrow x^3+\frac{1}{x^3}=343-21$

$\Rightarrow x^3+\frac{1}{x^3}=322$