anonymous
  • anonymous
Ok one more 16s^6-2g^6 so 2s(2s)^2-2g(g)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
hmmm....is that your simplification?
anonymous
  • anonymous
Yes am I wrong is this not the answer?
anonymous
  • anonymous
yeah 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

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anonymous
  • anonymous
Then am I to assume that (2s)^4 (2g)3
anonymous
  • anonymous
no because then you have 16s^4
anonymous
  • anonymous
one way to look at it is seperating the equation into a product of binomials
anonymous
  • anonymous
ok
anonymous
  • anonymous
hold on let me write something up first on paper to see if its correct
anonymous
  • anonymous
ok i got something
anonymous
  • anonymous
2s(2s)^5-2g(g)^5
anonymous
  • anonymous
so 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
  • anonymous
what you did is incorrect since (2s)^5 = 32s^5 and multiplying by 2s would be 64s^6
anonymous
  • anonymous
How about: \[-2(g^2-2s^2)(g^4+2g^2 s^2+4s^4)\]
anonymous
  • anonymous
dude if you look at what i wrote what you wrote is the exact same
anonymous
  • anonymous
Yes, after removing the open bracket symbol they are equivalent. Good work robotsandtoast.

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