Question

148) Factor as completely as possible with real coefficients: x^8-y^8

Answer 1

`(x^8-y^8)=(x^4+y^4)(x^2+y^2)(x-y)(x+y)`

Step-by-step explanation:

Here, we want to factor the polynomial completely

To do this, we will engage the use of the difference of two squares

We apply this rule to even exponents

We have it as follows:

`a^m-b^n=(a^{(m)/(2)}-b^{(n)/(2)})(a^{(m)/(2)}+b^{(n)/(2)})`

This is the law we are going to apply to factor the polynomial

We have this as follows:

`(x^8-y^8)=(x^4-y^4)(x^4+y^4)`

We can further split the even exponents with negative between them as follows:

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

Finally, we split the even exponent with negative as follows:

`(x^2-y^2)\text{ = (x+y)(x-y)}`

Thus, in cumulative, we have it as:

`(x^8-y^8)=(x^4+y^4)(x^2+y^2)(x-y)(x+y)`