To simplify the expression ( \sin^6 a + \cos^6 a + 3 \sin^2 a \cos^2 a ), we can use the identity for the sum of cubes and a bit of algebraic manipulation.
First, recall that:
[
x^6 + y^6 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)
]
Let ( x = \sin^2 a ) and ( y = \cos^2 a ).
Observe that:
[
\sin^2 a + \cos^2 a = 1
]
We can rewrite the expression using ( x ) and ( y ):
[
\sin^6 a + \cos^6 a = (\sin^2 a)^3 + (\cos^2 a)^3 = ( \sin^2 a + \cos^2 a ) ( \sin^4 a - \sin^2 a \cos^2 a + \cos^4 a )
]
This simplifies to:
[
1 \cdot ( \sin^4 a - \sin^2 a \cos^2 a + \cos^4 a )
]
Next, substitute ( \sin^4 a + \cos^4 a ) using the identity:
[
\sin^4 a + \cos^4 a = (\sin^2 a + \cos^2 a)^2 - 2\sin^2 a \cos^2 a = 1 - 2\sin^2 a \cos^2 a
]
Now we substitute this back into our expression:
[
\sin^6 a + \cos^6 a = (1 - 2\sin^2 a \cos^2 a) - \sin^2 a \cos^2 a = 1 - 3\sin^2 a \cos^2 a
]
Thus, we have:
[
\sin^6 a + \cos^6 a + 3\sin^2 a \cos^2 a = (1 - 3\sin^2 a \cos^2 a) + 3\sin^2 a \cos^2 a
]
This simplifies to:
[
1
]
Finally, we conclude that:
[
\sin^6 a + \cos^6 a + 3\sin^2 a \cos^2 a = 1
]
This confirms that the expression holds true, and thus the identity is verified.
