one of the factors of 64x^3+27y^3
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Rewrite 27y3 27 y 3 as (3y)3 ( 3 y ) 3 . Since both terms are perfect cubes, factor using the difference of cubes formula, a3−b3=(a−b)(a2+ab+b2) a 3 - b 3 = ( a - b ) ( a 2 + a b + b 2 ) where a=4x a = 4 x and b=3y b = 3 y . Simplify. Multiply 3 3 by −1 - 1 .
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