Question

Find the value of a:

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

Answer 1

Solution :- given

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

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

$\Rightarrow a^3+\frac{1}{3}=[a^3+\frac{1}{a^3}+3.a.\frac{1}{a}(a+\frac{1}{a})-3(a+\frac{1}{a})]$

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

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

$\Rightarrow a^3=3$

$\Rightarrow a={}^3\sqrt{3}$ (Ans)