Question

Let $n\in R$, such that; $n^{\frac{1}{3}}+\frac{1}{n^{\frac{1}{3}}}=3$. Then, $(n+\frac{1}{n},n^3+\frac{1}{n^3})=?$

• A. $(9,2778)$
• B. $(9,1800)$
• C. $(18,2700)$
• D. $(18,5778)$

Answer 1

The correct option is D $(18,5778)$

$n^{\frac{1}{3}}+\frac{1}{n^{\frac{1}{3}}}+(-3)=0$

$\Rightarrow a+b+c=0$, where $a=n^{\frac{1}{3}},b=\frac{1}{n^{\frac{1}{3}}},c=-3$

$\Rightarrow a^3+b^3+c^3-3abc=0$. (from the identity given in passage)

$\Rightarrow n+\frac{1}{n}+(-3{)}^3-3(n{)}^{\frac{1}{3}}{\left(\frac{1}{n}\right)}^{\frac{1}{3}}(-3)=0$
ie., $n+\frac{1}{n}=18$ ie., $n+\frac{1}{n}-18=0$

Now, $A+B+C=0$, where $A=n,B=\frac{1}{n},C=-18$

$\Rightarrow A^3+B^3+C^3-3ABC=0$ (since $A+B+C=0$)

$\Rightarrow (n{)}^3+{\left(\frac{1}{n}\right)}^3+(-18{)}^3=3\left(n\right)\left(\frac{1}{n}\right)(-18)$
ie., $n^3+\frac{1}{n^3}=5778$