Question

\[ \sin^6A+\cos^6A+3\sin^2A\cos^2A= \]

(a) \(0\)
(b) \(1\)
(c) \(2\)
(d) \(3\)

Solution

Using identity

\[ a^3+b^3=(a+b)^3-3ab(a+b) \]

Take

\[ a=\sin^2A,\quad b=\cos^2A \]

\[ \sin^6A+\cos^6A = (\sin^2A+\cos^2A)^3 -3\sin^2A\cos^2A(\sin^2A+\cos^2A) \]

\[ = 1^3-3\sin^2A\cos^2A(1) \]

\[ = 1-3\sin^2A\cos^2A \]

Therefore,

\[ \sin^6A+\cos^6A+3\sin^2A\cos^2A \]

\[ =1 \]

Answer

\[ \boxed{1} \]

Correct Option: (b)