Factoring y^3 + x + x^3 + y

Slecker

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I'm stuck on how to completely factor y^3 + x + x^3 + y

This is what I have so far:

Since addition is associative we can rearrange the terms so y^3 and x^3 are next to each other

= y^3 + x^3 + x + y

We see more clearly that y^3 + x^3 forms a sum of two cubes and we can apply the formula: A^3 + B^3 = (A + B) (A^2 - AB + B^2)

= (y + x) (y^2 - xy + x^2) + x + y

This is where I'm stuck, I don't know what to do with the "extra" x + y.

I looked up the solution in the back of the book and it's (y + x)(y^2 -xy +x^2 +1), which is pretty close to what I have so far, so I know I'm on the right track.
 
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I'm stuck on how to completely factor y^3 + x + x^3 + y

This is what I have so far:

Since addition is associative we can rearrange the terms so y^3 and x^3 are next to each other

= y^3 + x^3 + x + y

We see more clearly that y^3 + x^3 forms a sum of two cubes and we can apply the formula: A^3 + B^3 = (A + B) (A^2 - AB + B^2)

= (y + x) (y^2 - xy + x^2) + x + y

This is where I'm stuck, I don't know what to do with the "extra" x + y.

I looked up the solution in the back of the book and it's (y + x)(y^2 -xy +x^2 +1), which is pretty close to what I have so far, so I know I'm on the right track.

What you essentially have so far (where you've done almost all of the work already) is:

\(\displaystyle (x+y)(x^2-xy+y^2)+(x+y)\)

Now, you have 2 terms with \(\displaystyle (x+y)\) as a factor, which allows you to factor the entire expression as:

\(\displaystyle (x+y)((x^2-xy+y^2)+1)\)

And this is equivalent to what your textbook cites. Does this make sense?
 
What you essentially have so far (where you've done almost all of the work already) is:

\(\displaystyle (x+y)(x^2-xy+y^2)+(x+y)\)

Now, you have 2 terms with \(\displaystyle (x+y)\) as a factor, which allows you to factor the entire expression as:

\(\displaystyle (x+y)((x^2-xy+y^2)+1)\)

And this is equivalent to what your textbook cites. Does this make sense?

Oh, yeah now I see it. Essentially (x+y) is the greatest common factor among the terms which allows us to factor it out. I didn't see it until you pointed it out because I'm not used to grouping together individual terms with parentheses to create new terms. In fact, I wasn't aware that was a valid technique until now. At least I know for the future. Thanks!
 
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