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.