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(hope it doesn't seem so weird), I'm looking for a general expanded form of $(x+y+z)^k, k\in\mathbb{N}$.

$k=1: x+y+z$

$k=2: x^2+y^2+z^2+2xy+2xz+2yz$

$k=3: x^3+y^3+z^3+3xy^2+3xz^2+3yz^2+3x^2y+3x^2z+3y^2z+6xyz$

$k=4: x^4+y^4+z^4+4xy^3+4x^3y+4xz^3+4x^3z+4yz^3 +4y^3z+6x^2y^2+6y^2z^2+6x^2z^2+12x^2yz+12xy^2z+12xyz^2$

The elements are obviously determined by combinations of their powers, whose sum is always $k$.  I just cannot find the algorithm for element's constants

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3 Answers 3

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Hint: $(x+y+z)^k \equiv[\color{red}x+\color{green}{(y+z)}]^k$.

Now use the binomial formula for $(\color{red}a+\color{green}b)^k$.

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  • $\begingroup$ For the specific question, I much prefer this answer to mine. $\endgroup$ Jun 17, 2014 at 14:45
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    $\begingroup$ @JpMcCarthy imgur.com/iOFLp $\endgroup$
    – beep-boop
    Jun 17, 2014 at 14:46
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Note that just as you can use Pascal's Triangle for binomials, you can use Pascal's Pyramid for trinomials. Otherwise, you can use the Multinomial Theorem as Jp McCarthy suggested

Pascals' Pyramid

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It is the Multinomial Theorem.

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