Factor Completely

[mathdad]

Factor completely.

x^6 + 2x^3 + 1

Let u = x^3

Let x^6 = (x^3)^2

u^2 + 2u + 1

(u + 1)(u + 1)

Back-substitute for u.

(x^3 + 1)(x^3 + 1)

Each factor of (x^3 + 1) is the sum of cubes. I must apply the sum of cubes TWICE to find the final factored form. True?

 

[MarkFL]

I would write:

[MATH]x^6+2x^3+1=\left(x^3+1\right)^2[/MATH]
And then apply the sum of cubes factorization just once.

 

[mathdad]

I would write:

[MATH]x^6+2x^3+1=\left(x^3+1\right)^2[/MATH]
And then apply the sum of cubes factorization just once.

Much better.