If x3+1×3=110, then what is the value of x+1x?

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

1. A
5
2. B
10
3. C
15
4. D
None

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### Solution:

$={a}^{3}+{b}^{3}+3\mathit{ab}\left(a+b\right)$
${\left(x+\frac{1}{x}\right)}^{3}={x}^{3}+{\left(\frac{1}{x}\right)}^{3}+3×x×\frac{1}{x}\left(x+\frac{1}{x}\right)={x}^{3}+\frac{1}{{x}^{3}}+3\left(x+\frac{1}{x}\right)$
Let y = $\left(x+\frac{1}{x}\right)$
$⇒{y}^{3}=110+3y$
$⇒{y}^{3}-3y-110=0$

$⇒{y}^{2}\left(y-5\right)+5{y}^{2}-25y+22y-110=0$
$⇒{y}^{2}\left(y-5\right)+5{y}^{}\left(y-5\right)+22y\left(y-5\right)=0$
$⇒$ ) = 0
$⇒y=5$ or y = $\frac{-5±3\sqrt{7}i}{2}$
$⇒x+\frac{1}{x}=5$
Hence, Option (1) is the correct answer.

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