Question:
sin6 A + cos6 A + 3 sin2 A cos2 A =
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
We have:
$\sin ^{6} A+\cos ^{6} A+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)$
$=\left(\sin ^{2} A\right)^{3}+\left(\cos ^{2} A\right)^{3}+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right) \times 1$
$=\left(\sin ^{2} A\right)^{3}+\left(\cos ^{2} A\right)^{3}+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)\left(\sin ^{2} A+\cos ^{2} A\right)$
$=\left(\sin ^{2} A+\cos ^{2} A\right)^{3}$
$=1^{3}=1$
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