A smaller cube with side b (depicted by dashe | vedclass

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(A) We choose the origin at point $P$. The coordinates of the vertex $O$ of the smaller cube are $(b, b, b)$.Let $\rho$ be the mass density of the cub

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$X^3-X^2-X-1=0$

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(A) We choose the origin at point $P$. The coordinates of the vertex $O$ of the smaller cube are $(b, b, b)$.Let $ho$ be the mass density of the cube.The mass of the large cube is $M_1 = ho a^3$ and its centre of mass is at $(\frac{a}{2}, \frac{a}{2}, \frac{a}{2})$.The mass of the removed smaller cube is $M_2 = ho b^3$ and its centre of mass is at $(\frac{b}{2}, \frac{b}{2}, \frac{b}{2})$.Using the principle of negative mass for the centre of mass of the remaining solid:$x_{CM} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2} = b$$\Rightarrow \frac{ho a^3 (a/2) - ho b^3 (b/2)}{ho a^3 - ho b^3} = b$$\Rightarrow \frac{a^4 - b^4}{2(a^3 - b^3)} = b$$\Rightarrow a^4 - b^4 = 2b(a^3 - b^3)$Dividing by $b^4$ and substituting $X = a/b$:$X^4 - 1 = 2(X^3 - 1)$$X^4 - 2X^3 + 1 = 0$Since $X 
eq 1$,we divide by $(X-1)$:$(X-1)(X^3 - X^2 - X - 1) = 0$Thus,$X^3 - X^2 - X - 1 = 0$.

$A$ 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 $P$. Let $X = a/b$. If the centre of mass of the remaining solid is at the vertex $O$ of the smaller cube,then $X$ satisfies:

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