Q.If \(a, b, c, d\) are consecutive terms in a geometric progression, prove that \((a^2 - b^2)(c^2 - b^2) = (b^2 - c^2)^2\).

Let \(\cfrac{a}{b} = \cfrac{b}{c} = \cfrac{c}{d} = k\) [\(k\) is a nonzero constant].
\(\therefore a = bk, b = ck, c = dk\)
\(\therefore b = dk \times k = dk^2, a = dK^2 \times k = dk^3\)

LHS \(= (a^2 - b^2)(c^2 - d^2)\)
\(= \{(dk^3)^2 - (dk^2)^2\} \{(dk)^2 - d^2\}\)
\(= \{d^2k^6 - d^2k^4\} \{d^2k^2 - d^2\}\)
\(= d^2k^4(k^2 - 1) \times d^2(k^2 - 1)\)
\(= d^4k^4(k^2 - 1)^2\)

RHS \(= (b^2 - c^2)^2\)
\(= \{(dk^2)^2 - (dk)^2\}^2\)
\(= \{d^2k^4 - d^2d^2\}^2\)
\(= \{d^2k^2(k^2 - 1)\}^2\)
\(= d^4k^4(k^2 - 1)^2\)

\(\therefore\) LHS = RHS [Proved].
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