sin^6 A + cos^6 A= 1 - 3sin^2 A.cos^2 A
Last edited by snigdha; February 21st 2010 at 08:07 AM. Reason: typing error!
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Hello, snigdha! Prove: . The left side is a sum of cubes. . . . . . . Add and subtract . . . . . . . . . . . . . . . . . .
Originally Posted by Soroban Hello, snigdha! The left side is a sum of cubes. \sin^4\!A - \sin^2\!A\cos^2\!A + \cos^4\!A)" alt="\text{Factor: }\;\underbrace{(\sin^2\!A + \cos^2\!A)}_{\text{This is 1}}\\sin^4\!A - \sin^2\!A\cos^2\!A + \cos^4\!A)" /> . . . . . . Add and subtract . . . . . . . . . . . . . . . . . . i got a better clue... we know that a^3+b^3 = (a+b)^3 - 3ab(a+b) so by this i proceeded with the problem...& got it proved... thanks anyway...!!
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