Prove that 1 - (sin^6 x + cos^6 x) =(3sin^2 x)(cos^2 x)

We have to prove that 1 - (sin^6 x + cos^6 x) = 3(sin x)^2
(cos x)^2

Let's start with the left hand
side.

1 - [(sin x)^6 + (cos
x)^6]

=> (sin x)^2 + (cos x)^2 - (sin x)^6 - (cos
x)^6

=> (sin x)^2 - (sin x)^6 + (cos x)^2 - (cos
x)^6

=> (sin x)^2(1 - (sin x)^4) + (cos x)^2(1 -
(cos x)^4)

=> (sin x)^2(1 + (sin x)^2)(1 - (sin
x)^2) + (cos x)^2(1 + (cos x)^2)(1 - (cos x)^2)

=>
(sin x^2)(1 + (sin x)^2)(cos x)^2 + (cos x)^2(1 + (cos x)^2)(sin
x^2)

=> (sin x^2)(cos x)^2(1 + (sin x^2) + 1 + (cos
x)^2)

=> (sin x^2)(cos x)^2(1 + 1 +
1)

=> 3(sin x^2)(cos
x)^2

which is the right hand
side.

This proves : 1 - (sin^6 x + cos^6 x) =
3(sin x)^2 (cos x)^2