Expand ${(x + y + z)}^{4}$ $(x+y+z)^4$ ?

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Expand ${(x + y + z)}^{4}$ $(x+y+z)^4$ ?

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Answer 1 · 2018-03-21T07:40:50

${(x + y)}^{4} = x^{4} + y^{4} + z^{4} + 4 x^{3} y + 4 x y^{3} + 4 y^{3} z + 4 y z^{3} + 4 z^{3} x + 4 z x^{3} + 6 x^{2} y^{2} + 6 y^{2} z^{2} + 6 z^{2} x^{2} + 12 x^{2} y z + 12 x y^{2} z + 12 x y z^{2}$ $(x+y)^4 = x^4+y^4+z^4+4x^3y+4xy^3+4y^3z+4yz^3+4z^3x+4zx^3+6x^2y^2+6y^2z^2+6z^2x^2+12x^2yz+12xy^2z+12xyz^2$

Explanation:

Note that:

${(a + b)}^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$ $(a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4$

So we can find the terms of ${(x + y + z)}^{4}$ $(x+y+z)^4$ that only involve $2$ $2$ of $x, y, z$ $x, y, z$ by combining the expansions of binomial powers, One way to see that is to think about setting each of $x, y, z$ $x, y, z$ to zero in turn and expanding the remaining binomial.

By symmetry, the remaining terms - involving all three variables - will take the form $k x^{2} y z + k x y^{2} z + k x y z^{2}$ $kx^2yz+kxy^2z+kxyz^2$ for some constant $k$ $k$ .

So we have:

${(x + y + z)}^{4} = x^{4} + y^{4} + z^{4} + 4 x^{3} y + 4 x y^{3} + 4 y^{3} z + 4 y z^{3} + 4 z^{3} x + 4 z x^{3} + 6 x^{2} y^{2} + 6 y^{2} z^{2} + 6 z^{2} x^{2} + k x^{2} y z + k x y^{2} z + k x y z^{2}$ $(x+y+z)^4 = x^4+y^4+z^4+4x^3y+4xy^3+4y^3z+4yz^3+4z^3x+4zx^3+6x^2y^2+6y^2z^2+6z^2x^2+kx^2yz+kxy^2z+kxyz^2$

Putting $x = y = z = 1$ $x=y=z=1$ we have:

$81 = 3^{4}$ $81 = 3^4$

$81 = {(1 + 1 + 1)}^{4}$ $color(white)(81) = (1+1+1)^4$

$81 = 1 + 1 + 1 + 4 + 4 + 4 + 4 + 4 + 4 + 6 + 6 + 6 + k + k + k$ $color(white)(81) = 1+1+1+4+4+4+4+4+4+6+6+6+k+k+k$

$81 = 3 (1) + 6 (4) + 3 (6) + 3 k$ $color(white)(81) = 3(1)+6(4)+3(6)+3k$

$81 = 45 + 3 k$ $color(white)(81) = 45+3k$

So we have:

$3 k = 81 - 45 = 36$ $3k = 81-45 = 36$

So $k = 12$ $k = 12$ and:

${(x + y)}^{4} = x^{4} + y^{4} + z^{4} + 4 x^{3} y + 4 x y^{3} + 4 y^{3} z + 4 y z^{3} + 4 z^{3} x + 4 z x^{3} + 6 x^{2} y^{2} + 6 y^{2} z^{2} + 6 z^{2} x^{2} + 12 x^{2} y z + 12 x y^{2} z + 12 x y z^{2}$ $(x+y)^4 = x^4+y^4+z^4+4x^3y+4xy^3+4y^3z+4yz^3+4z^3x+4zx^3+6x^2y^2+6y^2z^2+6z^2x^2+12x^2yz+12xy^2z+12xyz^2$

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