find the value of x^3+y^3+12xy-64 when x+y=4
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(a+b)^3=a^3+b^3+3ab(a+b)
(x+y)^3=x3 +y3+3xy(x+y)
64=x3 +y3+3xy(4)
64=x3 +y3+12xy
64-12xy=x3 +y3
12xy-64+64-12xy=0
(x+y)^3=x3 +y3+3xy(x+y)
64=x3 +y3+3xy(4)
64=x3 +y3+12xy
64-12xy=x3 +y3
12xy-64+64-12xy=0
dhruvsaran1234:
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thank u bro....
you can go one more answer
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f8nd the integral zeroes of 2x^3+5x^2-5x-1
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