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Find the product of $$x^3+2x^2y+2xy^2+y^3 \quad \textrm{and} \quad x^3-2x^2y+2xy^2-y^3$$

I did a straightforward calculation of multiplying every term and got the correct answer $x^6-y^6$. My question is the two expressions are identical except having different signs. Is there a short-cut or insight that can simplify this multiplication without the need to multiply out every single term? Thank you.

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  • $\begingroup$ Well you actually have $$\left[ x^2 \left( x + 2y \right) + y^2 \left( 2x + y \right) \right] \left[ x^2 \left( x - 2y \right) + y^2 \left( 2x - y \right) \right]$$ Which is really just a lot of differences of two squares. This is one faster way. $\endgroup$
    – Bumblebee
    Commented Apr 4, 2023 at 10:08
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    $\begingroup$ $x^3-2x^2y+2xy^2-y^3 = (x-y)(x^2+xy+y^2)$ and $x^3+2x^2y+2xy^2+y^3 = (x+y)(x^2+xy+y^2)$ so multiplying them you get $(x^3-y^3)(x^3+y^3) = x^6 - y^6$ $\endgroup$
    – quantinho
    Commented Apr 4, 2023 at 10:38
  • $\begingroup$ You can make this a little easier by first considering what terms can appear in the final answer and computing the coefficients for each one. For example, in this problem we have all sixth degree terms, $x^6, x^5y,x^4y^2, x^3y^3, x^2y^4, xy^5, y^6$, as possible terms in the answer. We can compute the coefficients for each one without expanding the entire product and gathering terms. It's a little less error prone. $\endgroup$
    – John Douma
    Commented Apr 4, 2023 at 10:45

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We can rewrite each expression as $$x^3+y^3+2x^2y+2xy^2=(x+y)(x^2+xy+y^2)$$ $$x^3-y^3-2x^2y+2xy^2=(x-y)(x^2-xy+y^2)$$ Multiply both and pair them as $$\color{red}{(x+y)(x^2-xy+y^2)}\color{green}{(x-y)(x^2+xy+y^2)}$$ $$=\color{red}{(x^3+y^3)}\color{green}{(x^3-y^3)}$$ $$=\boxed{x^6-y^6}$$

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