Question

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

Answer 1

Answer:

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

Step-by-step 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:

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