# Finding real cubic root of the equation

The cubic equation has one real root.Find it. $\displaystyle 8x^3-3x^2-3x-1=0$

We have $$9x^3=(x+1)^3\iff \left(1+\frac1x\right)^3=9$$
Observe that $x$ will be real or complex according as $\displaystyle 1+\frac1x$
• I think that should be $\left(1+\dfrac1x\right)^3 = 9$. Jan 17 '14 at 12:05