How do you solve sin^6x+cos^6x=1-3sin^2xcos^2x?

Oct 27, 2016

It is not an equation that can be solved, but rather it is an identity.

Explanation:

Just to help simplify things let $S = {\sin}^{2} x$

Using the fundamental identity ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ we have;
$S + {\cos}^{2} x = 1 \implies {\cos}^{2} x = 1 - S$

So we can then write the initial equation:
${\sin}^{6} x + {\cos}^{6} x = 1 - 3 {\sin}^{2} x {\cos}^{2}$ as:

${S}^{3} + {\left(1 - S\right)}^{3} = 1 - 3 S \left(1 - S\right)$
$\therefore {S}^{3} + 1 - 3 S + 3 {S}^{2} - {S}^{3} = 1 - 3 S + 3 {S}^{2}$
$\therefore 0 = 0$

What does this mean! well it means the initial equation is true for all vales of x. Hence it is not an equation that can be solved, but rather it is an identity.

Oct 28, 2016

I think , It is an identity to prove.

$L H S = {\sin}^{6} x + {\cos}^{6} x$

$= {\left({\sin}^{2} x\right)}^{3} + {\left({\cos}^{2} x\right)}^{3}$

$= {\left({\sin}^{2} x + {\cos}^{2} x\right)}^{3} - 3 {\sin}^{2} x {\cos}^{2} x \left({\sin}^{2} x + {\cos}^{2} x\right)$

$= {\left(1\right)}^{3} - 3 {\sin}^{2} x {\cos}^{2} x \cdot \left(1\right)$

$= 1 - 3 {\sin}^{2} x {\cos}^{2} x = R H S$

Proved