How do you multiply #[x+(y+1)]^3#?

1 Answer
Aug 4, 2015

Answer:

Make the problem symmetrical by evaluating #(x+y+z)^3# then substitute #z = 1# to find:

#(x+(y+1))^3 = x^3+y^3+1+3x^2y+3y^2+3x+3x^2+3y^2x+3y+6xy#

Explanation:

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

The coefficient of #x^3# in the product is the number of ways of choosing a term from each of the trinomials so their product is #x^3#. This can only be done one way, by choosing #x# from each of the three trinomials. So the coefficient of #x^3# is #1#. By symmetry, the coefficients of #y^3# and #z^3# are also #1#.

The coefficient of #x^2y# in the product is the number of ways of choosing a term from each of the trinomials so their product is #x^2y#. We can choose one of the #3# trinomials to take #y# from, then choose #x# from the other two trinomials. This can be done in #3# ways, so the coefficient of #x^2y# in the product is #3#. By symmetry, #3# is also the coefficient of #y^2z#, #z^2x#, #x^2z#, #y^2x# and #zx^2#.

The coefficient of #xyz# in the product is the number of ways of choosing a term from each of the trinomials so their product is #xyz#. There are #3# ways to choose a trinomial to take #x# from. Then there are #2# remaining trinomials to pick #y# from. That leaves one trinomial to take #z# from. So the number of ways of forming #xyz# is #3xx2xx1 = 6#.

So

#(x+y+z)^3 = x^3+y^3+z^3+3x^2y+3y^2z+3z^2x+3x^2z+3y^2x+3z^2y+6xyz#

Now let #z=1# to get:

#(x+(y+1))^3 = x^3+y^3+1+3x^2y+3y^2+3x+3x^2+3y^2x+3y+6xy#