The Trinomial Theorem

Recall from The Binomial Theorem page that for all $x, y \in \mathbb{R}$ and $n \in \{ 0, 1, 2, ... \}$ we have that the expansion of the binomial $(x + y)^n$ is given by the formula:

(1)

\begin{align} \quad (x + y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1} x^{n-1}y^1 + ... + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n} x^0y^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} \end{align}

We saw in the Trinomial Coefficients page that the binomial coefficients $\binom{n}{k}$ could be rewritten if we let $r_1$ and $r_2$ be nonnegative integers where $r_1 = k$ and $r_2 = n - k$ so that $n = r_1 + r_2$ then $\binom{n}{k} = \binom{n}{r_1, r_2}$. Therefore, the binomial could therefore be rewritten as:

(2)

\begin{align} \quad (x + y)^n = \sum_{n = r_1 + r_2} \binom{n}{r_1, r_2} x^{r_1} y^{r_2} \end{align}

We will see that for the expansion of a trinomial $(x + y + z)^n$, an analogous theorem holds.

Theorem 1 (The Trinomial Theorem): If $x, y, z \in \mathbb{R}$, $r_1$, $r_2$, and $r_3$ are nonnegative integer such that $n = r_1 + r_2 + r_3$ then the expansion of the trinomial $(x + y + z)^n$ is given by $\displaystyle{(x + y + z)^n = \sum_{r_1 + r_2 + r_3 = n} \binom{n}{r_1, r_2, r_3} x^{r_1} y^{r_2} z^{r_3}}$.

• Proof: Let $x, y, z \in \mathbb{R}$. Consider the expansion of the trinomial $(x + y + z)^n$:

(3)

\begin{align} \quad (x + y + z)^n = \underbrace{(x + y + z) (x + y + z)...(x + y + z)}_{n \: \mathrm{factors}} \end{align}

• For each factor we choose to distribute through one of the three variables: $x$, $y$ or $z$. $r_1$ many times we choose to expand through $x$, $r_2$ many times we choose to expand through $y$, and $r_3$ many times we choose to expand through $z$ so that $n = r_1 + r_2 + r_3$.

• We therefore collect $\binom{n}{r_1, r_2, r_3}$ many (this coefficient tells us the number of ways that out of the $n$ factors we choose $r_1$ many $x$s, then out of the remaining $n - r_1$ factors we choose $r_2$ many $y$s, and out of the remaining $n - r_1 - r_2 = r_3$ factors we choose $r_3$ many $z$s) of the terms $x^{r_1} y^{r_2} z^{r_3}$ for each solution to $n = r_1 + r_2 + r_3$ where $r_1$, $r_2$, and $r_3$ are nonnegative integers. Therefore:

(4)

\begin{align} \quad (x + y + z)^n = \sum_{r_1 + r_2 + r_3 = n} \binom{n}{r_1, r_2, r_3} x^{r_1} y^{r_2} z^{r_3} \quad \blacksquare \end{align}

The only potential difficulty that arises in applying the Trinomial Theorem is finding all solutions nonnegative integer solutions to the equation $n = r_1 + r_2 + r_3$… but wait! We at least know how many of these solutions exist from the Nonnegative Integral Solutions to Simple Equations page.

For $r_1$, $r_2$, and $r_3$ as nonnegative integers we will have that there is exactly $\binom{3 + n - 1}{n}$ solutions to the equation $r_1 + r_2 + r_3 = n$.

For example, suppose that we want to expand the trinomial $(x + y + z)^3$. We will have there be $\binom{3 + 3 - 1}{3} = \binom{5}{3} = 10$ nonnegative integer solutions to this equation. They are the ordered pairs $(r_1, r_2, r_3)$ given in the table below:

$(3, 0, 0)$	$(0, 3, 0)$	$(0, 0, 3)$	$(2, 1, 0)$	$(2, 0, 1)$
$(0, 2, 1)$	$(1, 2, 0)$	$(0, 1, 2)$	$(1, 0, 2)$	$(1, 1, 1)$

Therefore by the Trinomial Theorem we have that:

(5)

\begin{align} \quad (x + y + z)^3 = x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2z + 3xy^2 + 3yz^2 + xz^2 + 6xyz \end{align}