Ask Me Help Desk - Binomial theorem

This is a trinomial expansion.

Using a Pascal-type triangle.

$\;\ \;\ \;\ \;\ \;\ \;\ x^{3} \;\ \;\ \;\ \;\$

$\;\ \;\ \;\ 3x^{2}y \;\ \;\ \;\ 3x^{2}z$

$\;\ 3xy^{2} \;\ \;\ \;6xyz \;\ \;\ \;\ \;\ 3xz^{2}$

$y^{3} \;\ \;\ 3y^{2}z \;\ \;\ \;\ 3yz^{2} \;\ \;\ \;\ \;\ z^{3}$

Just enter your values for x,y, and z.

There are some cool properties of the trinomial:

Reading down the left we have $(x+y)^{3}$

Reading down the right edge we have $(x+z)^{3}$

Reading across the bottom we have $(y+z)^{3}$

If we stack these up, we have Pascal's tetrahedron.

Every number is the sum of the 3 numbers above it.

i.e. At the top we have $x^{3}$ . Plug in $2x^{2}$ and get

$(2x^{2})^{3}=8x^{6}$

That is the first term starting at the left.

The next term would be at the $3x^{2}y$

Sub in $2x^{2}$ for x and $-4x$ for the y.

We get $-48x^{5}$ . That is the next term in the expansion. Continue.