Question

If $\frac{x+1}{x}=4$. Find $x^3+\frac{1}{x^3}$

Answer 1

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

Squaring (1)

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

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

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

Now $x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-x\times \frac{1}{x})$

$x^3+\frac{1}{x^3}=4\times (14-1)=4\times 13=52$

Hence the answer is 52.