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

$q:$ $x=0$.

To show that the above statement is true using direct method,

Let us assume that $p$ is true.

We have to show that $q$ is true.

$x$ is a real number such that $x^3+4x=0$

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

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

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

$\therefore$ $x=0$

$\therefore$ The given statement is true.