Question

Prove That :- cos^6 0 + sin^6 0 = 1 - 3 sin^2 o cos^2 0​

Answer 1

Given Question Correct Statement :-

$\tt \:  Prove \: that \: {cos}^{6} x + {sin}^{6} x = 1 - 3 {sin}^{2} x \: {cos}^{2} x$

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Identities used :-

$\boxed{\tt \:  1. \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)}$

$\boxed{\tt \:  2. \: {sin}^{2} x + {cos}^{2} x = 1}$

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$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  ⟼ \: Consider \: LHS$

$\tt \:  ⟼ \: {cos}^{6} x + {sin}^{6} x$

$\tt \:  = \: {( {cos}^{2}x) }^{3} + {( {sin}^{2}x) }^{3}$

$\tt \:  = {\bigg( {cos}^{2} x + {sin}^{2}x \bigg)}^{3} - 3 {cos}^{2} x \: {sin}^{2} x( {cos}^{2} x + {sin}^{2} x)$

$\tt \:  = {(1)}^{3} - 3 {sin}^{2} x {cos}^{2} x(1)$

$\tt \:  = 1 \: - \: 3 \: {cos}^{2} x \: {sin}^{2} x$

$\tt\implies \:➦ LHS = RHS$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$