factor 64^3-8y^3?

Question

my answer is (8x^2-4y^2) (8x+2y) am i right or wrong?

Answer 1

did you mean to type 64x^3 - 8y^3 ?
this is a difference of two cubes
(4x)^3 - (2y)^3
and factors this way (4x - 2y)(16x^2 + 8xy + 4y^2)

Answer 2

If you notice, the first and second terms are both perfect cubes. So you want to think back to the formula for the difference of perfect cubes, which is:

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

But before we do anything, we need to make sure we first take out any common factor. And we do have a GCF here since 8 can go into both terms. So we'll factor the 8 out first, and get:

8(8x^3 - y^3)

Okay, if you notice we still have a difference of cubes in our parentheses so we can go ahead and use our formula. So a = 2x and b = y:

8(2x - y)(4x^2 + 2xy + y^2) - This is our final answer!

Hope that made sense for you! =)


FYI - I'm assuming you meant 64x^3 - 8y^3 is the original problem even though you wrote 64^3 - 8y^3.....just wanted to clarify that.

Answer 3

This is a difference of CUBES, not squares! To see this, write

(4)^3 - (2y)^3. Use formula for diff of cubes,

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

(4-2y)(4^2 + 4*2y + (2y)^2) or
(4-2y)(16 + 8y + 4y^2)

Please note, the other answerers are wrong on this.

Answer 4

64^3 - 8y^3
64^3 - (2y)^3
(64 - 2y)(64^2 + 64*2y + (2y)^2)

Answer 5

64^3=8y^3.

8^3=y^3.
so y=8. you are wrong.

Answer 6

(4+2y)(16-8y +y^2)

Answer 7

64x^3-8y^3
(4x)^3-(2y)^3
this is of the form a^3-b^3
a^3-b^3=(a-b)(a^2+ab+b^2)
here a=4x & b=2y
a^3-b^3=(a-b)(a^2+ab+b^2)
(4x)^3-(2y)^3=(4x-2y)({4x}^2+{4x*2y}+{2y}^2)
=(4x-2y)(16x^2+8xy+4y^2)

Answer 8

64(x-.5y)(x^2 +.5xy+.25y^2)