Question

Ok one more
16s^6-2g^6
so
2s(2s)^2-2g(g)^2

Answer 1

hmmm....is that your simplification?

Answer 2

Yes am I wrong is this not the answer?

Answer 3

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

Answer 4

Then am I to assume that (2s)^4 (2g)3

Answer 5

no because then you have 16s^4

Answer 6

one way to look at it is seperating the equation into a product of binomials

Answer 7

ok

Answer 8

hold on let me write something up first on paper to see if its correct

Answer 9

ok i got something

Answer 10

2s(2s)^5-2g(g)^5

Answer 11

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 \[2[ (2s^{2} - g^{2})* ( (2s^{2})^{2} +2s^{2}g^{2} + (g^{2})^{2})\]

Answer 12

what you did is incorrect since (2s)^5 = 32s^5 and multiplying by 2s would be 64s^6

Answer 13

How about:
\[-2(g^2-2s^2)(g^4+2g^2 s^2+4s^4)\]

Answer 14

dude if you look at what i wrote what you wrote is the exact same

Answer 15

Yes, after removing the open bracket symbol they are equivalent. Good work robotsandtoast.