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anonymous
 5 years ago
Ok one more
16s^62g^6
so
2s(2s)^22g(g)^2
anonymous
 5 years ago
Ok one more 16s^62g^6 so 2s(2s)^22g(g)^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm....is that your simplification?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes am I wrong is this not the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah you have to notice that (2s)^2 = 4s^2 and so multiplying that by 2s would be 8s^3 and 2g*g^2 = 2g^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then am I to assume that (2s)^4 (2g)3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no because then you have 16s^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one way to look at it is seperating the equation into a product of binomials

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on let me write something up first on paper to see if its correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so notice factoring out a 2 you get \[2( 8s ^{6}  g^{6})\] and doing a little algebra we get \[2( (2s^{2})^{3}  (g^{2})^{3})\] and so using the difference of cubes we get \[2[ (2s^{2}  g^{2})* ( (2s^{2})^{2} +2s^{2}g^{2} + (g^{2})^{2})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what you did is incorrect since (2s)^5 = 32s^5 and multiplying by 2s would be 64s^6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How about: \[2(g^22s^2)(g^4+2g^2 s^2+4s^4)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dude if you look at what i wrote what you wrote is the exact same

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, after removing the open bracket symbol they are equivalent. Good work robotsandtoast.
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