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A smaller cube with side b (depicted by dashed lines) is excised from a bigger uniform cube with side a as shown below such that both cubes have a common vertex $\mathrm{P}$. Let $\mathrm{X}=\mathrm{a} / \mathrm{b}$. If the centre of mass of the remaining solid is at the vertex $\mathrm{O}$ of smaller cube then $\mathrm{X}$ satisfies.
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Verified Answer
The correct answer is:
$\mathrm{X}^{3}-\mathrm{X}^{2}-\mathrm{X}-1=0$
Centre of mass of remaining cube $x$ coordinate $=b$
$$
\mathrm{X}_{\mathrm{CM}}=\frac{\rho \mathrm{a}^{3} \times \frac{\mathrm{a}}{2}-\rho \mathrm{b}^{3} \times \frac{\mathrm{b}}{2}}{\rho \mathrm{a}^{3}-\rho \mathrm{b}^{3}}
$$
We will consider removed mass as a negative mass
$$
\begin{array}{l}
b=\frac{\frac{\rho a^{4}}{2}-\frac{\rho b^{4}}{2}}{\rho a^{3}-\rho b^{3}} \\
a^{3} b-b^{4}=\frac{a^{4}}{2}-\frac{b^{4}}{2} \\
2 a^{3} b-2 b^{4}=a^{4}-b^{4} \\
\text { put a }=b x \Rightarrow 2 b^{4} x^{3} b-2 b^{4}=b^{4} x^{4}-b^{4} \\
2 x^{3}-1==x^{4} \\
2 x^{3}-2+1=x^{4} \\
2\left[x^{3}-1\right]=\left(x^{2}-1\right)\left(x^{2}+1\right) \\
2[x-1]\left[x^{2}+1+x\right]=[x-1][x+1]\left[x^{2}+1\right] \\
2 x^{2}+2+2 x=x^{3}+x+x^{2}+1 \\
x^{3}-x^{2}-x-1=0
\end{array}
$$
$$
\mathrm{X}_{\mathrm{CM}}=\frac{\rho \mathrm{a}^{3} \times \frac{\mathrm{a}}{2}-\rho \mathrm{b}^{3} \times \frac{\mathrm{b}}{2}}{\rho \mathrm{a}^{3}-\rho \mathrm{b}^{3}}
$$
We will consider removed mass as a negative mass
$$
\begin{array}{l}
b=\frac{\frac{\rho a^{4}}{2}-\frac{\rho b^{4}}{2}}{\rho a^{3}-\rho b^{3}} \\
a^{3} b-b^{4}=\frac{a^{4}}{2}-\frac{b^{4}}{2} \\
2 a^{3} b-2 b^{4}=a^{4}-b^{4} \\
\text { put a }=b x \Rightarrow 2 b^{4} x^{3} b-2 b^{4}=b^{4} x^{4}-b^{4} \\
2 x^{3}-1==x^{4} \\
2 x^{3}-2+1=x^{4} \\
2\left[x^{3}-1\right]=\left(x^{2}-1\right)\left(x^{2}+1\right) \\
2[x-1]\left[x^{2}+1+x\right]=[x-1][x+1]\left[x^{2}+1\right] \\
2 x^{2}+2+2 x=x^{3}+x+x^{2}+1 \\
x^{3}-x^{2}-x-1=0
\end{array}
$$
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