<PRE><B>Question 519343</B>
(3x+3y)^2*(2x+3y)
First FOIL
(3x + 3y)* (3x + 3y) = 9x^2 + 9xy + 9xy + 9y^2
Combine like terms and you have
9x^2 + 18xy + 9y^2
:
Multiply this by (2x+3y), but do it this way, long multiplication
9x^2 + 18xy + 9y^2
. . . . . . . . . 2x + 3y 
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:
You should get: 18x^3 + 63x^y + 72xy^2 + 27y^3 
Corrected to: 18x^3  
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