Question

Prove that

sin^​6 A + cos^{​​​​6} A = 1 - 3 sin^{​​​​2} A cos^{​​​​2} A

Answer 1

Consider LHS:
sin6A+cos6A = (sin2A)3 + (cos2A)3
= (sin2A + cos2A)3 − 3 sin2Acos2A(sin2A + cos2A) [Since, (a+b)3 = a3 + b3+3ab(a+b)]
= (1)3 − 3 sin2A cos2A(1) [Since, sin2A+ cos2A = 1]
= 1 − 3 sin2A cos2A
= RHS