sin^6a + cos^6a + 3sin^2a ×cos^2a=1

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.