factorize:
cos^6A +sin^6A
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Answer:
cos^6A +sin^6A
=(cos²)³A+(sin²)³A
=(cosA+sinA)[(cos²A)²-cos²Asin²A+(sin²A)²]
=(cosA+sinA)[(cos²A)²+(sin²)²-cos²Asin²A]
=(cosA+sinA)[(cos²A+sin²A)²-2cos²Asin²A -cos²Asin²A]
=(cosA+sinA)[(1)²-3cos²Asin²A]
=(cosA+sinA)(1 - 3cos²Asin²A) .
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