The curves are y = x³, y = x + 6 and x = 0

On solving y = x³ and y = x + 6, we have

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

{{x}^{3}}~=\text{ }x\text{ }+\text{ }6  \\

{{x}^{3}}~\text{ }-x\text{ }-\text{ }6\text{ }=\text{ }0  \\

{{x}^{2}}\left( x\text{ }-\text{ }2 \right)\text{ }+\text{ }2x\left( x\text{ }-\text{ }2 \right)\text{ }+\text{ }3\left( x\text{ }-\text{ }2 \right)\text{ }=\text{ }0  \\

\left( x\text{ }-\text{ }2 \right)\text{ }\left( {{x}^{2~}}+\text{ }2x\text{ }+\text{ }3 \right)\text{ }=\text{ }0  \\

\end{array}\]

It’s seen that x² + 2x + 3 = 0 has no real roots

So, x = 2 is the only root for the above equation.

Now, the required area of the shaded region is given by