Question

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

• A. $6$
• B. $7$
• C. $8$
• D. $10$

Answer 1

Answer: (C)

Given, $x=1+\sqrt{2}$.

Now,

${(x-\frac{1}{x})}^3={(1+\sqrt{2}-(\frac{1}{1+\sqrt{2}}\left)\right)}^3$

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

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

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

$=(1+\sqrt{2}-\sqrt{2}+1{)}^3=2^3=8$.

Therefore, option $C$ is correct.