If sinx+cosx=a,

Given: $\sin x+\cos x=a$

Squaring on both sides, we get

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

$\Rightarrow 1+2 \sin x \cos x=a^{2}$

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

Now,

$\sin ^{6} x+\cos ^{6} x$

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

$=1-3\left(\frac{a^{2}-1}{2}\right)^{2} \quad[\operatorname{Using}(1)]$

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

Hence, the required value is $\frac{1}{4}\left[4-3\left(a^{2}-1\right)^{2}\right]$.