The given statement is

" If $x$ is real number such that $x^3+4x=0$, then $x$ is 0."

Let $p :$ If $x$ is real number such that $x^3+4x=0.$

and

$q:$ $x=0$.

$\Rightarrow\:$ ~$p:$ $x^3+4x\neq 0$ $\forall\:x \in R$. and

~$q$: $x\neq 0$.

To show that the above statement is true using contra positive method,

Let us assume that $q$ is false.

We have to prove that $p$ is also false.

$i.e.,$ Let $x\neq 0$ and $ x\in R$

$x^3+4x=0$

$\Rightarrow\:x(x^2+4)=0$

$\Rightarrow\:$ either $x=0$ or $x^2+4=0$

But our assumption is $x\neq 0$

$\Rightarrow\: x^2+4=0$

$x^2+4$ cannot be $0$ since $x^2 \geq 0\:\: \forall x \in R$

$\therefore$ $x^3+4x$ cannot be equal to zero

$\therefore$ The given statement is true.