We choose origin at point $P$.
Now, given centre of mass of remaining solid is at point $O$. Coordinates of $O$ as per axes choosen are $O \equiv(b, b, b)$
i.e. $x=b, y=b$ and $z=b$
Centre of mass of complete large cube lies at its centre.
$\therefore$ Coordinates of centre of mass of large cube are $x_1=\frac{a}{2}, y_1=\frac{a}{2}, z_1=\frac{ a }{2}$
And centre of mass of removed cube of side $b$ is $x_2=\frac{b}{2}, y_2=\frac{b}{2}, z_2=\frac{b}{2}$.
Treating removed mass as negative mass, we have
$x_{ CM }\left(\right.$ of remaining part) $=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}$
$\Rightarrow \quad b=\frac{\rho a^3 \times \frac{a}{2}-\rho b^3 \times \frac{b}{2}}{\rho a^3-\rho b^3}$
where, $\rho=$ mass density.
$\Rightarrow \quad a^3 b-b^4=\frac{a^4}{2}-\frac{b^4}{2}$
$\Rightarrow \quad \frac{a^3}{b^3}-1=\frac{a^4}{2 b^4}-\frac{1}{2}$
As $X=\frac{ a }{b}$, we have
$X^{-3}-1=\frac{X^4}{2}-\frac{1}{2}$
$\Rightarrow \quad 2 X^3-1=X^4$
$\Rightarrow \quad 2\left(X^3-1\right)=X^4-1$
$\Rightarrow 2(X-1)\left(X^2+1+X\right)$
$=(X-1)(X+1)\left(X^2+1\right)$
$\Rightarrow \quad X^3-X^2-X-1=0$
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