If sin x + cos x = a,

Given sin x + cos x = a

Squaring both sides, (sin x + cos x)2 = a2

$\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=a^{2}$

i. e $1+2 \sin x \cos x=a^{2}$

i. e $2 \sin x \cos x=a^{2}-1$     ...(1)

Using identity, we have

$a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$

$\sin ^{6} x+\cos ^{6} x=\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}$

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

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

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

$=1-3\left(\frac{a^{2}-1}{2}\right)^{2}$  (from (1))

$=1-\frac{3}{4}\left(a^{2}-1\right)^{2}=1-\frac{3}{4}\left(a^{4}+12 a^{2}\right)$

$=1-\left[\frac{3}{4} a^{4}+\frac{3}{4}-\frac{3}{2} a^{2}\right]$

$=1-\frac{3}{4}-\frac{3}{4} a^{4}+\frac{3}{2} a^{2}$

$=\frac{1}{4}=\frac{3}{4} a^{4}+\frac{3}{2} a^{2}$

Hence, sin6x + cos6x