Question

State true or false:

If $a+\frac{1}{a}=p$ and $a\ne 0$; then

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

• A. True
• B. False

Answer 1

The correct option is A True

$a+\frac{1}{a}=p$
Squaring both sides
$=>(a+\frac{1}{a}{)}^2=p^2$
$=>a^2+2\left(a\right)(\frac{1}{a}+(\frac{1}{a}{)}^2=p^2$
$=>a^2+2+\frac{1}{a^2}=p^2$
$=>a^2+\frac{1}{a^2}=p^2-2$
Now,
$a^3+\frac{1}{a^3}=(a+\frac{1}{a})(a^2-a(\frac{1}{a})+(\frac{1}{a^2}\left)\right)$
$=>a^3+\frac{1}{a^3}=\left(p\right)(p^2-2-1)$
$=>a^3+\frac{1}{a^3}=\left(p\right)(p^2-3)$
$=>L.H.S=R.H.S$