Help me factor 64x*6+y*6?

Question

could you please show me by step by step i hate factoring

its an exponent

Answer 1

(y^2+4*x^2)(16*x^4-4*x^2*y^2+y^4)

Answer 2

Do you mean 64x^6 + y^6? (That is, with the 6's as exponents?)

If so, this is the sum of two cubes.
a^3+b^3=(a+b)(a^2-ab+b^2), so
(4x^2)^3 + (y^2)^3 = (4x^2+y^2)(16x^4-4x^2y^2+y^4)

Let's play with that second factor some. When I see 16x^4 and y^4, I want a middle term of 8x^2y^2. I say this because
(4x^2+y^2)^2=16x^4+8x^2y^2+y^4.

However, 8x^2y^2 is not our middle term; instead, we have -4x^2y^2. That's not bad news; watch:

16x^4 - 4x^2y^2 + y^4 = 16x^4 + [8x^2y^2 - 12x^2y^2] + y^4
= [16x^4 + 8x^2y^2 + y^4] - 12x^2y^2
= (4x^2+y^2)^2 - 12(xy)^2, the difference of two squares
...and Sqrt(12)=2Sqrt(3)...
= [4x^2 + y^2 + 2xySqrt(3)][4x^2 + y^2 - 2xySqrt(3)]

I don't believe any of those can be factored further, so I'm left with the final answer

64x^6 + y^6 = (4x^2 + y^2)[4x^2 + y^2 + 2xySqrt(3)][4x^2 + y^2 - 2xySqrt(3)].

Answer 3

balerina, does * mean ^ power?

Aaah! Then see: f(x,y)=64x^6+y^6; Let 2x =u, y=u·v; then f(x,y)= u^6 (1+v^6);
Assume v^6 = -1, while –1= exp((2k +1)·j·π) and k= -2,-1, 0, 1,2,3; thence v[k] = exp((2k +1)·j·π/6); Now consider 3 pairs in {…}:

By the way do you know complex numbers? If yes I continue!

Doc B has correct answer, so i quit!
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Answer 4

6(64x+y)

Answer 5

Assuming that by x*6 you mean "x raised to the 6th power"... normally online that's represented by x^6.

What you have is of the form a²+b² = (a+bi)(a-bi)

(8x³ +y³ i)(8x³ -y³ i)