Let \[p:\]’In case \[x\]is a genuine number to such an extent that \[{{x}^{3}}\text{ }+\text{ }4x\text{ }=\text{ }0,\]then, at that point, \[x\text{ }is\text{ }0’\]

\[q:\text{ }x\]is a genuine number to such an extent that \[{{x}^{3}}\text{ }+\text{ }4x\text{ }=\text{ }0\]

\[r:\text{ }x\text{ }is\text{ }0\]

By contrapositive strategy, to show proclamation \[p\]to be valid, we expect that \[r\]is bogus and prove that \[q\]should be bougus

\[\sim r:~x~\ne \text{ }0\]

Unmistakably, it tends to be seen that

\[({{x}^{2}}~+\text{ }4)~\]will consistently be positive

\[x\text{ }\ne \text{ }0\] infers that the result of any sure genuine number with \[x\]isn’t zero.

Presently, consider the result of \[x\]with \[({{x}^{2}}~+\text{ }4)\]

\[\begin{array}{*{35}{l}}

\therefore ~x~\left( {{x}^{2}}~+\text{ }4 \right)\text{ }\ne \text{ }0  \\

{{x}^{3}}~+\text{ }4x\text{ }\ne \text{ }0  \\

\end{array}\]

This shows that assertion \[q\]isn’t correct.

Thus, demonstrated that

\[\sim r\Rightarrow \sim q\]

Thus, the given assertion \[p\]is valid.