Question

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Show that :-
$\sf \: {sin}^{8} \theta - {cos}^{8} \theta = (1 - 2 {cos}^{2} \theta)(1 - 2 {sin}^{2} \theta \: {cos}^{3} \theta)$

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Answer 1

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$\dashrightarrow \: \: \bf \: {sin}^{8} (\theta) - {cos}^{8} (\theta )$

$\dashrightarrow \: \: \sf \: \big({ \sin {}^{4} ( \theta) } \big)^{2} - \big({ \cos{}^{4} ( \theta) } \big)^{2}$

$\dashrightarrow \: \: \sf \big({ \sin {}^{4} ( \theta) } + { \cos {}^{4} ( \theta) } \big) \big({ \sin {}^{4} ( \theta) } - { \cos {}^{4} ( \theta) } \big)$

$\dashrightarrow \: \: \sf \bigg(\big({ \sin {}^{2} ( \theta) } + { \cos {}^{2} ( \theta) } \big) {}^{2} - 2 { \sin^{2}( \theta) } \cos {}^{2} ( \theta) \bigg)\bigg({ \sin {}^{2} ( \theta) } + { \cos {}^{2} ( \theta) } \bigg)\bigg({ \sin {}^{2} ( \theta) } - { \cos {}^{2} ( \theta) } \bigg)$

$\dashrightarrow \: \: \sf \bigg(1 - 2 { \sin^{2}( \theta) } \cos {}^{2} ( \theta) \bigg)\bigg(1 - { \cos{}^{2} ( \theta) } - { \cos {}^{2} ( \theta) } \bigg)$

$\dashrightarrow \: \blue{\frak{ \bigg(1 - 2 { \sin^{2}( \theta) } \cos {}^{2} ( \theta) \bigg)\bigg(1 - 2 { \cos {}^{2} ( \theta) } \bigg) }} \: \: \: \: \: \: \: \: \: \bf \: \: proved.$

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Answer 2

LHS = RHS

(Hence Proved)

Step-by-step explanation:

Refer attachment.