<s>[INST] How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
 [/INST] This is a cubic equation. The forme of cubic equation is $x^3+px+q=0$ where 
$1)$ $p$, $q$ $\in$ R
$2)$ $D=(\frac{q}{2})^2+(\frac{p}{3})^3$
$3)$ $x=u+v=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2 +(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2 +(\frac{p}{3})^3}}$ 
For the given equation we have:
$p=-12$, $q=16$, $D=(\frac{16}{2})^2+(\frac{-12}{3})^3=8^2+(-4)^3=64-64=0$.
Because $D=0$, $u=v=\sqrt[3]{\frac{-q}{2}}$ $\Rightarrow$ $u=v=\sqrt[3]{\frac{-16}{2}}=\sqrt[3]{-8}$ $\Rightarrow$  $u_1=v_1=-2$.
Definitly 
$x_1=2u_1$ $\Rightarrow$ $x_1=-4$
$x_2=x_3=-u_1$ $\Rightarrow$ $x_2=x_3=2$
</s>