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 smaller cube then X satisfies.

Step-by-step solution

Centre of mass of remaining cube x coordinate =b X _ CM = a ³ a 2 - b ³ b 2 a ³- b ³ We will consider removed mass as a negative mass array l b= a⁴ 2 - b⁴ 2 a³- b³ a³ b-b⁴= a⁴ 2 - b⁴ 2 2 a³ b-2 b⁴=a⁴-b⁴ put a =b x 2 b⁴ x³ b-2 b⁴=b⁴ x⁴-b⁴ 2 x³-1==x⁴ 2 x³-2+1=x⁴ 2 [x³-1 ]= (x²-1 ) (x²+1 ) 2[x-1] [x²+1+x ]=[x-1][x+1] [x²+1 ] 2 x²+2+2 x=x³+x+x²+1 x³-x²-x-1=0 array