Prove the following identities: `2(sin^6theta+cos^6theta)-3(sin^4theta+cos^4theta)+1=0` `sin^6theta+cos^6theta+3sin^2thetacos^2theta=1` `(sin^8theta-cos^8theta)=(sin^2theta-cos^2theta)(1-2s in^2thetacos^2theta)`

Explanation

To prove the given identities step by step, let's start with the first identity:

### Identity 1:
**Prove that:**
\[ 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 = 0 \]

#### Step 1: Rewrite \(\sin^6 \theta + \cos^6 \theta\)
Using the identity for the sum of cubes, we can express \(\sin^6 \theta + \cos^6 \theta\) as:
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