Question

State true or false:

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

$a^3+\frac{1}{a^3}=0$

• A. True
• B. False

Answer 1

The correct option is A True

${(a+\frac{1}{a})}^2=3$
$(a+\frac{1}{a}{)}^2=3$
$=>a^2+\frac{1}{a^2}+2\left(a\right)\left(\frac{1}{a}\right)=3$
$=>a^2+\frac{1}{a^2}=3-2$
$=>a^2+\frac{1}{a^2}=1$
And
$=>(a+\frac{1}{a})=\sqrt{3}$
Now,
$=>a^3+\frac{1}{a^3}=(a+\frac{1}{a})\left[\right(a^2)+(\frac{1}{a^2})-(a\left)\right(\frac{1}{a}\left)\right]$
$=\sqrt{3}(1-1)$
$=0$
Therefore,
L.H.S=R.H.S