Question

If $a+\frac{1}{a}=p$ and $a\ne 0$; then show that:
$a^3+\frac{1}{a^3}=p(p^2-3)$

Answer 1

$a+\frac{1}{a}=p$

Squaring both sides
${(a+\frac{1}{a})}^2=p^2$

$a^2+2\left(a\right)\frac{1}{a}+{\left(\frac{1}{a}\right)}^2=p^2$

$a^2+2+\frac{1}{a^2}=p^2$

$a^2+\frac{1}{a^2}=p^2-2$

Now,
$a^3+\frac{1}{a^3}=(a+\frac{1}{a})(a^2-a\left(\frac{1}{a}\right)+\left(\frac{1}{a^2}\right))$

$a^3+\frac{1}{a^3}=\left(p\right)(p^2-2-1)$

$a^3+\frac{1}{a^3}=\left(p\right)(p^2-3)$

$L.H.S=R.H.S$