Question

Expand `(x+y+z)^4`?

Answer 1

`(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+4a^3b+6a^2b^2+4ab^3+b^4`

So we can find the terms of `(x+y+z)^4` that only involve `2` of `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` to zero in turn and expanding the remaining binomial.

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

So we have:

`(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` we have:

`81 = 3^4`

`color(white)(81) = (1+1+1)^4`

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

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

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

So we have:

`3k = 81-45 = 36`

So `k = 12` and:

`(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`