Question

If $a^2+\frac{1}{a^2}=34$, then the value of $a^3+\frac{1}{a^3}$ is equal to.

• A. $\pm 234$
• B. $98$
• C. $\pm 198$
• D. $\pm 99$

Answer 1

The correct option is C $\pm 198$

It is given that
$a^2+\frac{1}{a^2}=34$
Using identity we have
$(a+\frac{1}{a}{)}^2=a^2+\frac{1}{a^2}+2$
$\Rightarrow$ $(a+\frac{1}{a}{)}^2=34+2$
$\Rightarrow$ $(a+\frac{1}{a}{)}^2=36$
$\Rightarrow$ $(a+\frac{1}{a}{)}^2=6^2$
$\Rightarrow$ $(a+\frac{1}{a})=\pm 6$

Now, again from cubic identity we have,
$a^3+\frac{1}{a^3}=(a+\frac{1}{a}{)}^3-3(a\left)\right(\frac{1}{a}\left)\right(a+\frac{1}{a})$

Case I: When $(a+\frac{1}{a})$ = +6,
$a^3+\frac{1}{a^3}=6^3-3\left(a\right)\left(\frac{1}{a}\right)6$
$\Rightarrow$ $a^3+\frac{1}{a^3}=6^3-\left(3\right)\left(6\right)$
$\Rightarrow$ $a^3+\frac{1}{a^3}=216-18$
$\Rightarrow$ $a^3+\frac{1}{a^3}=+198$

Case II: When $(a+\frac{1}{a})$ = -6,
$a^3+\frac{1}{a^3}=(-6{)}^3-3(a\left)\right(\frac{1}{a}\left)\right(-6)$
$\Rightarrow$ $a^3+\frac{1}{a^3}=(-6{)}^3-(3\left)\right(-6)$
$\Rightarrow$ $a^3+\frac{1}{a^3}=-216+18$
$\Rightarrow$ $a^3+\frac{1}{a^3}=-198$

Hence,
$a^3+\frac{1}{a^3}=\pm 198$