How to expand 3 variables using Pascal's triangle?

#(x+y+z)^3#

1 Answer
Dec 22, 2017

It does not help much, but...

Explanation:

The instant response to this question might be to say that Pascal's triangle does not help, since it is concerned with powers of binomials. For example, the row #1, 3, 3, 1# in Pascal's triangle helps us find that:

#(x+y)^3 = x^3+3x^2y+3xy^2+y^3#

What we can do is to combine the results of applying Pascal's triangle as follows:

If #x = 0# then:

#(x+y+z)^3 = (y+z)^3 = y^3+3y^2z+3yz^2+z^3#

If #y = 0# then:

#(x+y+z)^3 = (z+x)^3 = z^3+3z^2x+3zx^2+x^3#

If #z = 0# then:

#(x+y+z)^3 = (x+y)^3 = x^3+3x^2y+3xy^2+y^3#

The three expressions on the right hand side are essentially the full expression we are looking for, with some terms missing. The first one is missing any terms involving #x#, the second any terms involving #y#, the third any terms involving #z#. So if none of #x, y, z# is necessarily zero then mushing these together we must have:

#(x+y+z)^3 = x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2+kxyz#

for some #k# to be determined.

Note that the sum of the coefficients must be #3^3 = 27# so:

#k = 27-(1+1+1+3+3+3+3+3+3) = 6#