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anonymous
 4 years ago
Can someone please show me step by step, how to factor this equation? a^3xb^3xa^3y+b^3y
the answer = (ab)(a^2+ab+b^2)(xy)
anonymous
 4 years ago
Can someone please show me step by step, how to factor this equation? a^3xb^3xa^3y+b^3y the answer = (ab)(a^2+ab+b^2)(xy)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a^3xb^3xa^3y+b^3y (a^3b^3)x  (a^3 b^3)y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay, yes. First you take the x and y out

Hezekieli
 4 years ago
Best ResponseYou've already chosen the best response.0One could also start by taking a and b out:\[a^{3}(xy)+b^{3}(yx)\] But I don't really know how to get to that answer or why to aim at that? :o

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The best way to do it: FIND THE COMMON NUMBER BETWEEN EVERYTHING.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a^3b^3=(ab)(a^2+ab+b^2)\]

Hezekieli
 4 years ago
Best ResponseYou've already chosen the best response.0You can check that what abdul said by going backwards: \[(a  b)(a ^{2} + ab + b ^{2})\] \[= (a^{3}+a^{2}b + ab^{2}) + (a^{2}b  ab^{2}  b^{3})\] \[= a^{3}+a^{2}b  a^{2}b + ab^{2}  ab^{2}  b^{3}\] \[= a^{3}  b^{3}\] Not sure yet how to get that (x  y) out o.O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a^3xb^3xa^3y+b^3y\] \[x(a^3b^3)y(a^3b^3)\] \[(a^3b^3)(xy)\] \[(ab)(a^2+ab+b^2)(xy)\]
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