If \(a, b, c\) are in continued geometric progression, then \(a : b = b : c\). Let \(\frac{a}{b} = \frac{b}{c} = k\), where \(k \ne 0\). So, \(a = bk = ck^2\) and \(b = ck\) --- **Left-hand side (LHS):** \[ a^2 b^2 c^2 \left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}\right) = (ck^2)^2 \cdot (ck)^2 \cdot c^2 \left(\frac{1}{(ck^2)^3} + \frac{1}{(ck)^3} + \frac{1}{c^3}\right) \] \[ = c^2k^4 \cdot c^2k^2 \cdot c^2 \left(\frac{1}{c^3k^6} + \frac{1}{c^3k^3} + \frac{1}{c^3}\right) = c^6k^6 \cdot \frac{1}{c^3} \left(\frac{1}{k^6} + \frac{1}{k^3} + 1\right) \] \[ = c^3k^6 \left(\frac{1 + k^3 + k^6}{k^6}\right) = c^3(k^6 + k^3 + 1) \] --- **Right-hand side (RHS):** \[ a^3 + b^3 + c^3 = (ck^2)^3 + (ck)^3 + c^3 = c^3k^6 + c^3k^3 + c^3 = c^3(k^6 + k^3 + 1) \] --- ∴ LHS = RHS ( proved)