factorise 27 x^3 +y^3
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17
Answer:
Rewrite 27x3 27 x 3 as (3x)3 ( 3 x ) 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=3x a = 3 x and b=y . Simplify.
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7
Answer:
Rewrite 27x3 27 x 3 as (3x)3 ( 3 x ) 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=3x a = 3 x and b=y .
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