+118609
\(x+\frac1x=-5, \;\;\;what \;\;\;is\;\;\;x^3+\dfrac1{x^3}\\~\\ \left(x+\frac{1}{x}\right)^3=x^3+3x^2*\frac{1}{x}+3x*\frac{1}{x^2}+\frac{1}{x^3}\\ \left(x+\frac{1}{x}\right)^3=x^3+3x+\frac{3}{x}+\frac{1}{x^3}\\ \left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})\\ sub\\ (-5)^3=x^3+\frac{1}{x^3}+3(-5)\\ -125=x^3+\frac{1}{x^3}-15\\ -110=x^3+\frac{1}{x^3}\\ x^3+\frac{1}{x^3}=110 \)
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+610 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 $$x^3+\frac1{x^3}=(-125)-(-15)=\boxed{-110}.$$
