How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
 A: Can you see by inspection that 2 is a root? If so, we can write your equation as $x^{3}-12x+16=(x-2)(ax^2+bx+c)=ax^{3}+(b-2a)x^{2}+(c-2b)x-2c$ by the factor theorem.
Thus, $a=1$, $b-2a=0$, $c-2b=-12 $ and $-2c=16$
Which gives $a=1$, $b=2$ and $c=-8$.
Therefore $x^{3}-12x+16=(x-2)(x^2+2x-8)=(x-2)(x-2)(x+4)=(x-2)^{2}(x+4)$
So there is a repeated root at $x=2$ and a second at $x=-4$.
A: 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 
$x_1=2u_1$ $\Rightarrow$ $x_1=-4$
$x_2=x_3=-u_1$ $\Rightarrow$ $x_2=x_3=2$
A: Hint: $x=2$ is a solution of the polynomial.
So $x^3-12x+16$ is divisible by $(x-2)$.
And $x^3-12x+16= (x-2)(x^2+2x-8)$.
The roots of $x^3-12x+16=0$ are $x=2$ and the roots of $x^2+2x-8=0$.
A: HINT: 

$$
\begin{align*}
x^3-12x-16&=x\underbrace{(x^2-4)}_{(x-2)(x+2)}-8x-16 \tag{1}\\
&=x(x-2)(x+2)-\underbrace{8x-16}_{-8(x+2)}\\
&=(x+2) \underbrace{(x^2-2x-8)}_{\text{quadratic equation}}=0 
\\
\end{align*}$$

Now just solve the quadratic equation and you should get something like:
$$(x+2)(x+a)(x+b)=0\tag{2}$$
and find:  $\,\,x=-2,x=-a \,\,\text{or}\,\,x=-b$
A: HINT: Since $$(s-t)^3+3st(s-t)+s^3-t^3=0$$ Thus, if you could find $s,t$ such that $s^3-t^3=16$ and $3st=-12$ then $x=s-t$ is a root.
EDIT: Taking $3st=-12\implies t=-\frac{4}{s}$ and putting it int equation $s^3-t^3=16\implies s^3-(\frac{-64}{s^3})=16$ Now let $y=s^3$ which converts it into quadratic equation in $y$ and you can solve it and then $s=y^{1/3}$. after getting $s$, compute corresponding $t$ and then $x=s-t$ is a root of the given equation.
In this particular equation $x=2$ satisfies the equation. Now to find other two roots divide $x^3-12x+16$ by $x-2$ to get a quadratic equation and solve it to get other roots 
