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 contradiction method,
Let us assume that $p$ is true.
and $q$ is not true.
$i.e.,$ $x$ is a real number such that $x^3+4x=0$ and $x\neq 0$
$\Rightarrow\:\:x(x^2+4)=0$
$\Rightarrow\:$ either $x=0$ or $x^2+4=0$
But our assumption is $x\neq 0$ and
also $x^2+4$ cannot be $0$ since $x^2 \geq 0\:\: \forall x \in R$
$\therefore$ $x^3+4x$ cannot be equal to zero which is contradiction.
$\therefore$ $x$ should be equal to $0$.
$\therefore$ The given statement is true.