Question

If $(y+\frac{1}{y})=12$, then $(y^3+\frac{1}{y^3})$ is equal to:-

• A. $1680$
• B. $1686$
• C. $1692$
• D. $1698$

Answer 1

Answer: (C)

We have,

$(y+\frac{1}{y})=12$

On taking cube both sides, we get

${(y+\frac{1}{y})}^3={12}^3$

$y^3+\frac{1}{y^3}+3(y+\frac{1}{y})=1728$

$y^3+\frac{1}{y^3}+3\times 12=1728$

$y^3+\frac{1}{y^3}+36=1728$

$y^3+\frac{1}{y^3}=1692$

Hence, this is the answer.