Question

If $a+\frac{1}{a}=3$, then find the value of $(a^2+\frac{1}{a^2})(a^3+\frac{1}{a^3})$

• A. 123
• B. 125
• C. 126
• D. 112

Answer 1

The correct option is C

126

Given, $a+\frac{1}{a}=3$ ... (i)

Squaring both the sides, we get

${(a+\frac{1}{a})}^2=9$

$\Rightarrow a^2+\frac{1}{a^2}+2a.\frac{1}{a}=9$

$\Rightarrow a^2+\frac{1}{a^2}=9-2=7$ ... (ii)

Now, taking cube of equation (i), we get

${(a+\frac{1}{a})}^3=27$

$\Rightarrow a^3+\frac{1}{a^3}+3a.\frac{1}{a}(a+\frac{1}{a})=27$

$\Rightarrow a^3+\frac{1}{a^3}+3\times 3=27$

$\Rightarrow a^3+\frac{1}{a^3}=27-9=18$ ... (iii)

We have to find the value of $(a^2+\frac{1}{a^2})(a^3+\frac{1}{a^3})$

$=7\times 18=126$