If a, b, c are in G.P., then show that
i) $(a^{2} - b^{2}) (b^{2} + c^{2}) = (b^{2} - c^{2}) (a^{2} + b^{2})$
ii) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

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Solution

$a, b, c$ are in G.P
Let $r$ be common ratio
So $b = a r$ & $c = a r^{2}$
$(1) L . H . S (a^{2} - b^{2}) (b^{2} + c^{2}) = (a^{2} - a^{2} r^{2}) (a^{2} r^{2} + a^{2} r^{4}) = a^{2} (1 - r^{2}) a^{2} r^{2} (1 + r^{2}) = a^{4} r^{2} (1 - r^{4}) R . H . S (b^{2} - c^{2}) (a^{2} + b^{2}) = (a^{2} r^{2} - a^{2} r^{4}) (a^{2} + a^{2} r^{2}) = a^{2} r^{2} (1 - r^{2}) a^{2} (1 + r^{2}) = a^{4} r^{2} (1 - r^{4}) L . H . S = R . H . S p r o v e d$
$L . H . S a (b^{2} + c^{2}) = a (a^{2} r^{2} + a^{2} r^{4}) = a^{3} r^{2} (1 + r^{2}) R . H . S = c (a^{2} + b^{2}) = a r^{2} (a^{2} + a^{2} r^{2}) = a^{3} r^{2} (1 + r^{2}) L . H . S = R . H . S p r o v e d$

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