Question

If $x^3+\frac{1}{x^3}=110$ then the value of $x+\frac{1}{x}$ is

• A. $8$
• B. $7$
• C. $3$
• D. $5$

Answer 1

Answer: (D)

$5$

We know that one of the algebraic formula is ${(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$

It is given that $x^3+\frac{1}{x^3}=110$, therefore, using the above formula we have:

${(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})\Rightarrow x^3+\frac{1}{x^3}={(x+\frac{1}{x})}^3-3(x+\frac{1}{x})\Rightarrow {(x+\frac{1}{x})}^3-3(x+\frac{1}{x})=110(Given{x}^3+\frac{1}{x^3}=110)$

Let $x+\frac{1}{x}=k$, we get:

$k^3-3k=110\Rightarrow k(k^2-3)=110\Rightarrow k(k^2-3)=5(5^2-3)$

By comparing, we get $k=5$.

Hence, $x+\frac{1}{x}=5$.