Prove that :
2(sin6A+cos6A)—3(sin4A+cos4A)+1=0
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2 ( sin⁶ + cos⁶ ) - 3 ( sin⁴ + cos⁴ ) + 1
= 2 [( sin² + cos² )³ - 3 sin² cos²( sin² + cos² )] - 3 [( sin² + cos² )² - 2 sin² cos²] + 1
= 2 [ (1)³ - 3 sin² cos² (1) ] - 3 [ (1)² - 2 sin² cos² ] + 1
= 2 - 6 sin² cos² - 3 + 6 sin² cos² + 1
= 0
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