Question

If $x^2+\frac{1}{x^2}=7$ then find the value of $x^3+\frac{1}{x^3}$ using only the positive value of $x+\frac{1}{x}$.

Answer 1

Given: $x^2+\frac{1}{x^2}=7$

Adding $2$ on both the sides, we get,

$\Rightarrow x^2+\frac{1}{x^2}+2=7+2$

$\Rightarrow x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=9$

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

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

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

Cubing both the sides, we get,

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

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

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

$\Rightarrow x^3+\frac{1}{x^3}=27-9$

$\therefore x^3+\frac{1}{x^3}=18$