How do I factor this then?

Question

8r^6 + 27s^12

I don't see what there is to factor out. Would it be like

(2r^2+3s^4)(4r^4-6r^2s^4+9s^8) Or what?

Answer 1

Your answer is absolutely correct..congratulations. Those are not always easy for students because it's memorization of a pattern

Answer 2

Ok, let's try this one:

8r^6 = (2^3)(r^3)^2 = (2r^2)^3
27s^12 = (3^3)(s^4)^3 = (3s^4)^3

(2r^2)^3 + (3s^4)^3

now, you need to use the a^3 + b^3 identity
(a^2 - ab + b^2) (a+b)
let a= 2r^2 and b = 3s^4

(4r^4 - 6r^2s^4 + 9s^8) (2r^2 + 3s^4)

which is the result you just got from the answer book.

Good luck!

Answer 3

It the sum of cubes:
X 3+ Y 3 = (x + y)(X 2 - XY + Y 2 )

8r^6 + 27s^12 = (2r^2)^3 + (3s^4)^3
SO
(2r^2 + 3s^4)[(2r^2)^2 - (2r^2 x 3s^4) + (3s^4)^2]
= (2r^2 + 3s^4)(4r^4 - 6r^2s^4 + 9s^8)

Answer 4

this kind of equation is called the sum of cubes

generally, it will look like this

a^3+b^3

It could be factored like this:

(a+b)(a^2-ab+b^2)

In the first term, the cube root of a^3 and b^3 were just added.

The second term is based from the first term.

a in the first term were just squared to get a^2.

we got -ab because we multiply a and b of the first term then multiply it to -1.

b in the first term were just squared to get b^2.

in your case

8r^6+27s^12

by following the steps i gave you a while ago, you will get.

(2r^2+3s^4)(4r^4-6r^2s^4+9s^8)

Answer 5

The first parenthesis is correct, the second is this:
(4r^4 - 2*3*r^2*s^4 + 9*s^8) the same as you have.

Answer 6

(2r^2)^3+(3s^4)^3
(2r^2+3s^4)(4r^4-6r^2s^4+9s^8)