If sin +cos=a prove sin6+cos6=4-3(a²-1)²/4

LHS
{tex} = {\sin ^6}\theta + {\cos ^6}\theta {/tex}
{tex} = {\left( {{{\sin }^2}\theta } \right)^3} + {\left( {{{\cos }^2}\theta } \right)^3}{/tex}
{tex} = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^3} - 3{\sin ^2}\theta {\cos ^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right){/tex}
{tex}\left[ {\because {a^3} + {b^3} = {{(a + b)}^3} - 3ab(a + b)} \right]{/tex}
{tex} = {(1)^3} - 3{\sin ^2}\theta {\cos ^2}\theta \times 1{/tex} {tex}\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]{/tex}
{tex} = 1 - 3{\sin ^2}\theta {\cos ^2}\theta {/tex}
RHS {tex} = \frac{{4 - 3{{({x^2} - 1)}^2}}}{4}{/tex}
{tex} = \frac{{4 - 3{{\left\{ {{{\left( {\sin \theta + \cos \theta } \right)}^2} - 1} \right\}}^2}}}{4}{/tex} {tex}\left[ {given\;x = \sin \theta + \cos \theta } \right]{/tex}
{tex} = \frac{{4 - 3{{\left\{ {{{\sin }^2}\theta + {{\cos }^2}\theta + 2\sin \theta \cos \theta - 1} \right\}}^2}}}{4}{/tex}
{tex} = \frac{{4 - 3{{\left\{ {1 + 2\sin \theta \cos \theta - 1} \right\}}^2}}}{4}{/tex} {tex}\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]{/tex}
{tex} = \frac{{4 - 3{{\left( {2\sin \theta \cos \theta } \right)}^2}}}{4}{/tex}
{tex} = \frac{{4 - 3 \times 4{{\sin }^2}\theta {{\cos }^2}\theta }}{4}{/tex}
{tex} = \frac{{4\left( {1 - 3{{\sin }^2}\theta {{\cos }^2}\theta } \right)}}{4}{/tex}
LHS = RHS
Hence proved.