Question

Which expression is the completely factored form of x^6– 64y^3?​

Answer 1

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

Explanation:

Factoring

We need to recall the following polynomial identity:

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

The given expression is:

`x^6- 64y^3`

To factor the above expression, we need to find a and b, knowing a^3 and b^3. a is the cubic root of a^3, and b is the cubic root of b^3:

`a=\sqrt[3]{x^6}=x^2`

`b=\sqrt[3]{64y^3}=4y`

Now we apply the identity:

`x^6– 64y^3=(x^2-4y)[(x^2)^2+(x^2)*(4y)+(4y)^2]`

Operating:

`x^6– 64y^3=(x^2-4y)[x^4+4x^2y+16y^2]`

The remaining factors cannot be factored in anymore. Thus the completely factored form is:

`\boxed{(x^2-4y)(x^4+4x^2y+16y^2)}`