Prove that: `3(sintheta-costheta)^4+6(sintheta+costheta)^2+4(sin^6theta+cos^6theta)-13=0.`

Explanation

To prove that \( 3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) - 13 = 0 \), we will break down the expression into three parts and simplify each part step by step.

### Step 1: Simplifying \( 3(\sin \theta - \cos \theta)^4 \)

Let \( a = \sin \theta - \cos \theta \).

Using the identity \( (a - b)^2 = a^2 + b^2 - 2ab \):

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