Is 1-(sin^6x+cos^6x)=(3sin^2x)(cos^2x) a Valid Trigonometric Identity?

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The equation 1-(sin^6x+cos^6x)=(3sin^2x)(cos^2x) is a valid trigonometric identity. The manipulation begins with the left-hand side, transforming it into 1-(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x). This simplifies further to 1-(sin^4x-sin^2xcos^2x+cos^4x), which can be equated to the right-hand side (3sin^2x)(cos^2x). The discussion emphasizes algebraic manipulation as a key technique in proving trigonometric identities.

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larymac47
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1-(sin^6x+cos^6x)=(3sin^2x)(cos^2x)

I got this far:

1-(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)=(3sin^2x)(cos^2x)
1-(sin^4x-sin^2xcos^2x+cos^4x)=(3sin^2x)(cos^2x)
 
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I'd start by manipulating one of the sides.
 

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