Question

If $x^2+\frac{1}{x^2}=23$, find the value of $x^3+\frac{1}{x^3}$ if $x$ is a positive integer.

Answer 1

$x^2+\frac{1}{x^2}=23$

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

$x^2+\frac{1}{x^2}+2=25$

$(x+\frac{1}{x}{)}^2=5^2$

$\therefore x+\frac{1}{x}=5$ --- (1)

$(x+\frac{1}{x}{)}^3=5^3$

$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=125$

Put the value of (1)

$x^3+\frac{1}{x^3}+3\times 5=125$

$x^3+\frac{1}{x^3}=125-15=110$