Question 118725


{{{xy(x^2+3xy+9)}}} Start with the given expression



{{{(xy)(x^2)+(xy)(3xy)+(xy)(9)}}} Distribute {{{xy}}} among the terms in the parenthesis



{{{x^3y+(xy)(3xy)+(xy)(9)}}} Multiply {{{xy}}} and  {{{x^2}}} to get {{{x^3y}}}

 

{{{x^3y+3x^2y^2+(xy)(9)}}} Multiply {{{xy}}} and  {{{3xy}}} to get {{{3x^2y^2}}}

 

{{{x^3y+3x^2y^2+9xy}}} Multiply {{{xy}}} and  {{{9}}} to get {{{9xy}}}

 


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Answer:

So {{{xy(x^2+3xy+9)}}} distributes to {{{x^3y+3x^2y^2+9xy}}}


In other words, {{{xy(x^2+3xy+9)=x^3y+3x^2y^2+9xy}}}