Question

Can someone explain the factoring of the sum or difference of cubes

Answer 1

do you mean when you have (x^3 +8)??

Answer 2

like c^3 + D^3

Answer 3

the one I an having alot of trouble on is v^7 + 27v

Answer 4

i would just factor that by taking a 'v' out

Answer 5

then I have V^6 + 27 then when factoring I get (v^2)^3 + (3)^3

Answer 6

you would actually get v(v^6 + 27) that factoring method just seems to get you no where

Answer 7

I have too though

Answer 8

ok...i can actually see the point in it after i said that haha

Answer 9

then according to the answer key I would get v(c^2 +3){(V^2)^2 - 3V^2 + 3^2]

Answer 10

wtf...let me see the whole problem

Answer 11

http://www.wolframalpha.com/input/?i=factor+v^7+%2B+27v

Answer 12

the whole problem is this

\[v ^{7} + 27v\]\[ = v(v ^{6} + 27(\]

then

\[v[(v ^{2})^{3} + 3^{3}]\]

then

\[v(v ^{2} + 3)[(v ^{2)^{2 }} - 3v ^{2} + 3^{2}]\]

which all is finally = to

\[v(v ^{2} + 3)(v ^{4} - 3v ^{2} + 9)\]

But I can't figure out the freakin problem that is from the answer key

Answer 13

I get lost here v(v^2+3)[(v^2)^2−3v^2+3^2]

Answer 14

The sum of cubes factoring formula is


x^3+y^3 = (x+y)(x^2-xy+y^2)


In the case of v^6+27, which is really (v^2)^3+3^3, we see that


x = v^2 and y = 3



So plug them into the formula above to get



(v^2)^3+3^3 = (v^2+3)((v^2)^2-(v^2)(3)+(3)^2)


(v^2)^3+3^3 = (v^2+3)(v^4-3v^2+9)