Question
$\sin^6\text{A}+\cos^6\text{A}+3\sin^2\text{A}\cos^2\text{A}=$
- 0.
- 1.
- 2.
- 3.
Solution:
We have:
$\sin^6\text{A}+\cos^2\text{A}+3(\sin^2\text{A})(\cos^2\text{A})$
$=(\sin^2\text{A})^3+(\cos^2\text{A})^3+3(\sin^2\text{A})(\cos^2\text{A})\times1$
$=(\sin^2\text{A})^3+(\cos^2\text{A})^3+3(\sin^2\text{A})(\cos^2\text{A})(\sin^2\text{A}+\cos^2\text{A})$
$=(\sin^2\text{A}+\cos^2\text{A})^3$
$=1^3=1$
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