Question

If $a+\frac{1}{a}=4$ and $a\ne 0$, then find $a^3+\frac{1}{a^3}$

• A. $52$
• B. $50$
• C. $51$
• D. $53$

Answer 1

The correct option is A $52$

$a+\frac{1}{a}=4$

Squaring both sides,

$=(a+\frac{1}{a}{)}^2=4^2$

$=a^2+(\frac{1}{a}{)}^2+2(a\left)\right(\frac{1}{a})=16$

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

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

Now,

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

$=4(a^2-1+\frac{1}{a^2})$

$=4(14-1)$

$=4\left(13\right)$

$=52$