Question

Simplify
\[\frac{a^3-b^-3}{b-a}*frac{6}{2a^+2ab+b^2}\]

Answer 1

please re write the problem

Answer 2

\[\frac{a^3-b^3}{b-a}*\frac{6}{2a^2+2ab+b^2}\]
I got 3

Answer 3

k wait

Answer 4

Ugh, sorry correct \[\frac{a^3-b^3}{b-a}*\frac{6}{2a^2+2ab+2b^2}\]

Answer 5

the correct answer is -3

Answer 6

hold up

Answer 7

has to be a minus in the right side, since when you multiply out the right side you need things to cancel so there must be OPPOSITE signs in the first factor (a+b) and the second factor (a^2-ab+b^2) to make that cancellation happen.

Similarly, in a^3-b^3=(a-b)(a^2+ab+b^2), you have to have OPPOSITE signs in the two factors on the right side or there will be no cancellation. witch =3

Answer 8

u have to use integration here

Answer 9

yes yes

Answer 10

ok @truck I don't understand what you're saying... and @mathslover ingeration? This is for algebra I

Answer 11

\[\frac{a^3-b^3}{b-a}*\frac{6}{2a^2+2ab+2b^2}\]\[=\frac{(a-b)(a^2 + ab+b^2)}{b-a}*\frac{6}{2(a^2+ab+b^2)}\]\[=-1*3\]\[=-3\]