Question

Number of real solution of the equation $x^3+3x^2-(x-6{)}^{\frac{1}{3}}+8+3x=0$ is

Answer 1

$(x+1{)}^3=(x-6{)}^{\frac{1}{3}}-7$
$\begin{array}{cc}\underset{}{(x+1{)}^3+6}=& \underset{}{(x-6{)}^{\frac{1}{3}}-1}\\ f\left(x\right)& f^{-1}\left(x\right)\end{array}$
So solution is f(x) = x
$x^3+3x^2+3x+7=x$
$x^3+3x^2+2x+7=0$
$(x+1{)}^3=x-6$