Rewrite x^6 as (x^3)^2.

Let u = x^3

u^2 + 2u + 1

Factor.

(u + 1)(u + 1)

Back-substitute for u.

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

The sum of cubes tells me that (x^3 + 1) is the same thing as

(x + 1)(x^2 - x + 1).My answer is (x + 1)(x^2 - x + 1)^2.

Textbook answer is (x + 1)^2(x^2 - x + 1)^2.

Who is right and why?

in the red part you have

in the blue part you factored

then replace

the aquare goes on **both** factors inside

once you do that youll match the books answer

We're in luck here because (x^3)^2 = x^6

So substitute Y = x^3 and you'll get a quadratic in Y.

Bob

Let me see.

Rewrite x^6 as (x^3)^2.

Let u = x^3

u^2 + 2u + 1

Factor.

(u + 1)(u + 1)

Back-substitute for u.

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

The sum of cubes tells me that (x^3 + 1) is the same thing as

(x + 1)(x^2 - x + 1).

My answer is (x + 1)(x^2 - x + 1)^2.

Textbook answer is (x + 1)^2(x^2 - x + 1)^2.

Who is right and why?

]]>Factor completely. If the polynomial cannot be factored, say it is prime.

x^6 + 2x^3 + 1

Bob wrote:

We're in luck here because (x^3)^2 = x^6

harpazo1965 wrote:

You said we're in luck in this case because (x^3)^2 = x^6.

What if the question is not so obvious? What then?

then maybe it cant be factored, right?

if your doing perfect sqares then it has to fit into perfect aquare pattern

Thank you. You said we're in luck in this case because (x^3)^2 = x^6.

What if the question is not so obvious? What then?

So substitute Y = x^3 and you'll get a quadratic in Y.

Bob

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