If \(a, b, c\) are in continued geometric progression, then \(\cfrac{a}{b} = \cfrac{b}{c}\) Let \(\cfrac{a}{b} = \cfrac{b}{c} = k\), where \(k \ne 0\) \(\therefore a = bk = ck^2\), and \(b = ck\)
Left-hand side: \(\cfrac{1}{b^2} = \cfrac{1}{(ck)^2} = \cfrac{1}{c^2k^2}\)
Right-hand side: \(\cfrac{1}{b^2 - a^2} + \cfrac{1}{b^2 - c^2}\) \(= \cfrac{1}{(ck)^2 - (ck^2)^2} + \cfrac{1}{(ck)^2 - c^2}\) \(= \cfrac{1}{c^2k^2 - c^2k^4} + \cfrac{1}{c^2k^2 - c^2}\) \(= \cfrac{1}{c^2k^2(1 - k^2)} + \cfrac{1}{c^2(k^2 - 1)}\) \(= \cfrac{1}{c^2k^2(1 - k^2)} - \cfrac{1}{c^2(1 - k^2)}\) \(= \cfrac{1 - k^2}{c^2k^2(1 - k^2)} = \cfrac{1}{c^2k^2}\)
\(\therefore\) Left-hand side = Right-hand side (Proved)