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

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Solution

Consider the left hand side
LHS = ${sin}^{6} x + {cos}^{6} x$
= $({sin}^{2} x)^{3} + ({cos}^{2} x)^{3}$
= $({sin}^{2} x + {cos}^{2} x)^{3} - 3 {sin}^{2} x {cos}^{2} x ({sin}^{2} x + {cos}^{2} x)$
= $(1)^{3} - 3 {sin}^{2} x {cos}^{2} x \times 1$
= $1 - 3 {sin}^{2} x {cos}^{2} x$
=RHS
Hence proved.

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