+0

+1
49
2
+477

If $x+\frac1x=-5$, what is $x^3+\frac1{x^3}$?

gueesstt  Apr 9, 2018
Sort:

#1
+92206
+2

$$x+\frac1x=-5, \;\;\;what \;\;\;is\;\;\;x^3+\dfrac1{x^3}\\~\\ \left(x+\frac{1}{x}\right)^3=x^3+3x^2*\frac{1}{x}+3x*\frac{1}{x^2}+\frac{1}{x^3}\\ \left(x+\frac{1}{x}\right)^3=x^3+3x+\frac{3}{x}+\frac{1}{x^3}\\ \left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})\\ sub\\ (-5)^3=x^3+\frac{1}{x^3}+3(-5)\\ -125=x^3+\frac{1}{x^3}-15\\ -110=x^3+\frac{1}{x^3}\\ x^3+\frac{1}{x^3}=110$$

Melody  Apr 9, 2018
#2
+477
0

You were almost correct! It was actually -110. Thank you for the help:

To get $x^3$ and $\frac1{x^3}$, we cube $x+\frac1x$: $$-125=(-5)^3=\left(x+\frac1x\right)^3=x^3+3x+\frac3x+\frac1{x^3}$$ by the Binomial Theorem. Conveniently, we can evaluate $3x + \frac{3}{x}$ as $3\left(x+\frac1x\right)=3(-5)=-15$, so $$x^3+\frac1{x^3}=(-125)-(-15)=\boxed{-110}.$$

gueesstt  Apr 9, 2018

### 18 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details