The value of $$3(sin x- cos x)^4$$ + $$6(sin x + cos x)^2$$ + $$4(sin^{6}x + cos^{6}x)$$ is

Expression : $$3(sin x- cos x)^4$$ + $$6(sin x + cos x)^2$$ + $$4(sin^{6}x + cos^{6}x)$$

= $$3 (sin x - cos x)^2 (sin x - cos x)^2 + 6 (sin^2 x + cos^2 x + 2 . sin x . cos x) + 4 [(sin^2 x)^3 + (cos^2 x)^3]$$

= $$3 (sin^2 x + cos^2 x - 2 . sin x . cos x) ( sin^2 x + cos^2 x - 2 . sin x . cos x) + 6 (1 + 2 . sin x . cos x) + 4[(sin^2 x + cos^2 x) (sin^4 x + cos^4 x - sin^2 x . cos^2 x)]$$

= $$3 (1 - 2 . sin x . cos x)^2 + 6 + 12 . sin x . cos x + 4 [(sin^2 x + cos^2 x)^2 - 3 . sin^2 x . cos^2 x]$$

= $$3 (1 + 4 . sin^2 x . cos^2 x - 4 . sin x . cos x) + 6 + 12 . sin x . cos x + 4 - 12 . sin^2 x . cos^2 x$$

= $$3 + 12 . sin^2 x . cos^2 x - 12 . sin x . cos x + 6 + 12 . sin x . cos x + 4 - 12 . sin^2 x . cos^2 x$$

= $$3 + 6 + 4 = 13$$