Question

\(x^3-12x^2+48x-72=0\)

Answer 1

\(x^3-3.4.x^2+3.x.16-64-8=0\\ \left(x-4\right)^3-8=0\\ \left(x-4-2\right)\left[\left(x-4\right)^2+2\left(x-4\right)+4\right]=0\\ \left(x-6\right)\left(x^2-8x+16+2x-8+4\right)=0\\ \left(x-6\right)\left(x^2-6x+12\right)=0\\ \left[{}\begin{matrix}x-6=0\\x^2-6x+12=0\left(vono\right)\end{matrix}\right.=>x=6\)