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

$$(sin^{6}x + cos^{6}x) = ((sin^2)^3x + (cos^2)^3x)$$
=$$ ((sin^2x)+ (cos^2x))((sin^4x)+ (cos^4x)+ sin^2x cos^2x$$
=$$(sin^4)+ (cos^4) + 2sin^2xcos^2 - 3sin^2xcos^2x$$
=$$(sin^2x + cos^2x)^2 - 3sin^2xcos^2x$$
=$$1 - 3sin^2xcos^2x$$

$$(sinx+cosx)^{2} $$= $$1 + sinxcosx$$

$$(sinx -cos x)^4 = (1-2sinxcosx)^2$$
$$= 1 + 4sin^2xcos^2x - 4sinxcosx$$

$$3(sinx- cosx)^{4}+6(sinx+cosx)^{2}+4(sin^{6}x + cos^{6}x)=3( 1+ 4sin^2xcos^2x - 4sinxcosx) + 6(1 +2sinx cosx) +4(1 -3sin^2xcos^2x)$$
=$$3 + 12sin^2xcos^2x - 12sinxcosx + 6 + 12sinxcosx + 4 -12 sin^2xcos^2 x$$
=$$13$$
Option D is the correct answer.