how to factor this kind of eq?

4x^3-256=0 ---------- Add 256
4x^3=256 ---------- divide by 4
x^3 = 64 ----------- take the cubed root of each side
x = 4 (Since 4*4*4 = 64)

4X^3-256=0
OR 4(X^3- 64)=0
or x^3-64=0
or x^3-4^3=0
now use formula
a^3-b^3=(a-b)(a^2+ab+b^2)
x^3-4^3=(x-4)(x^2+4x+16)=0

4x^3 - 256 = 0
(4x^3 - 256)/4 = 0/4
x^3 - 64 = 0
x^3 - 4^3 = 0
(x - 4)(x^2 + 4x + 4^2) = 0
(x - 4)(x^2 + 4x + 16) = 0

x^2 + 4x + 16 = 0 (it must be imaginary number)

x - 4 = 0
x = 4

∴ x = 4

uhm, for me...

4x^3 - 256 = 0 transpose

4x^3 = 256 divide it by 4 which is the coefficient of x..
x^3 = 64.. cube root..
cube root of x^3 = cube root of 64..
x = 4..

First, notice that there is a common factor in this binomial; the common factor is 4. Factor it out:
4(x^3 - 64) = 0

So, you've got a difference of two cubes in the parentheses.
Difference of two cubes is of the form: (a^3 - b^3)
Its factors: (a - b)(a^2 + ab + b^2)
Here: a^3 = x^3; which means that a = x
b^3 = 64; which means that b = 4
Therefore, factoring 4(x^3 - 64) = 0 further will yield:

4(x - 4)(x^2 + 4x + 16) = 0 Answer

Take note that you stated "how to factor," and NOT "how to solve for x."

But if you are really solving for x, one solution is x = 4, since x - 4 = 0.

You may solve for the other value/s of x from the trinomial x^2 + 4x + 16 = 0 either by completing the square or the quadratic equation. Either way, the solution would be: x = -2 + or - 2*sqrt(-3)

4 x ³ = 256
x ³ = 64
x = 4

4x^3 = 256
x^3 = 256/4
= 64
x = 4

EDIT*
I admit defeat to h.b. g, he has the best answer out of all of us...

4(x^3 - 64) = 0

There's nothing special about this problem, don't get caught up by the exponent and large number. Four go into both 4*x^3 and 256 (which is 4*64), so you can factor four out.

if you wanted to solve this problem, just divide both sides by 4 to get
x^3 - 64 = 0
move 64 to the other side
x^3 = 64
Then get the cube root of both sides (64 is 4*4*4)
x = 4

Done

Haha "Im answering to see what other people answer" lol

Remember to choose a best answer!

Im answering to see what other people answer

Good Luck :D