Any Alg. experts that could show me how to factor this?

Question

16x^3 + 54y^6

I can see that you take a (2) out and then you have cubes (8x^3)+(27y^6), but then what do you do? Any experts out there?

LoL nice try Scarlet, but the guy below you had it done way before you ;-)

Answer 1

A^3 + B^3 = (A+B)(A^2 - AB + B^2)

Hence we have
16x^3 + 54y^6
= 2(8x^3 + 27y^6)
= 2[(2x)^3 + (3y^2)^3]
= 2(2x + 3y^2)(4x^2 - 6xy^2 + 9y^4)

Answer 2

16x^3 + 54y^6 = 2(8x^2 + 27y^6)

You may think this is it, but notice that the expression in the bracket is a sum of cubes. There is an identity:

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

Applying the identity, we get,

2(8x^3 + 27y^6) = 2[(2x + 3y^2)(4x^2 - 6xy^2 + 9y^4)]

Now, the expression is fully factored

Answer 3

Factoring out 2 leaves you with the
sum of 2 cubes 8x^3 + 27y^6

Answer:
2 [ (2x + 3y^2) (4x^2 - 6xy^2 + 9y^4) ]

Answer 4

2[8x^3 + 27y^6]
= 2 [(2x)^3 + (3y^2)^3]

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

= 2 (2x + 3y^2) (4x^2 - 6xy^2 + 9y^4)

Hope this helps.

Answer 5

Given exp
=2(8x^3+27y^6)
=2{(2x)^3+(3y^2)^3]
=2(2x+3y^2){(2x)^2-2x*3y^2
+(3y^2)^2
=2(2x+3y^2)(4x^2-6xy^2+9y^4)

Answer 6

a^3 + b^3 has a factor (a+b); it factors as (a+b) (a^2- ab+b^2).

So we have
16x^3 + 54y^6
= 2 ((2x)^3 + (3y^2)^3)
= 2 (2x + 3y^2) (4x^2 - 6xy^2 + 9y^4).

Answer 7

2.( 8.x^3 + 27.y^6)
= 2.( (2.x)^3 + (3.y^2)^3) [since y^6 = (y^2)^3]
= 2.(2.x + 3.y^2)(4.x^2 + 3.y^4 - 6.x.y^2)

[using d formula a^3 + b^3 = (a+b)(a^2 + b^2 - ab)]

Thats as far as i know, any further steps aren't possible

Answer 8

2 (2x + 3y^2)·(4x^2 - 6xy^2 + 9y^4)