Expanding A Cubic

Consider the expansion of $(x_1+x_2+\ldots+x_n)^3$ . There are terms of the form $x_i^3$ , terms of the form $3x_i^2 x_j$ (with $i \neq j$ ) and terms of the form $6x_i x_j x_k$ ( $i,j,k$ distinct). Respectively, there are $n$ , $n(n-1)$ , $\frac16 n(n-1)(n-2)$ of these terms; so we need to solve

$100n=n+n(n-1)+\frac16 n(n-1)(n-2)$

This simplifies to $n^2+3n-598=0$ , which has the unique positive root $n=\boxed{23}$ .

However, this is only guaranteed to work when the terms of the cubed polynomial don't share any variables. For instance, $(x+y+z)^3$ has $10$ terms in its expanded form, but $(x^2+x+1)^3$ only has $7$ . I'm not sure if this means there's another solution.

For $n \ge 3$ The only possible different terms of the cubed polynomial are $a^3$ , $ab^2$ , and $abc$ , where $a$ , $b$ , and $c$ are distinct terms of the polynomial $x_1, x_2, x_3, \cdots x_n$ . For $n$ -term polynomial:

• The number of $a^3$ terms is $N_{a^3} = n$
• The number of $a^2b$ terms is $N_{a^2b} = 2\dbinom n2$
• The number of $abc$ terms is $N_{abc} = \dbinom n3$

Therefore, the number of distinct terms in the cubed $n$ -term polynomial is:

$\begin{aligned} N & = N_{a^3} + N_{a^2b} + N_{abc} \\ & = n + 2\binom n2 + \binom n3 \\ & = n + 2 \left(\frac {n(n-1)}2\right) + \frac {n(n-1)(n-2)}6 \\ & = n \left(1+(n-1)\left(1+\frac {n-2}6\right)\right) \\ & = \frac {n(n+1)(n+2)}6 \end{aligned}$

For $N=100n$ , then:

$\begin{aligned} \frac {n(n+1)(n+2)}6 & = 100n \\ (n+1)(n+2) & = 600 = \left \lfloor \sqrt{600} \right \rfloor \left \lceil \sqrt{600} \right \rceil = 24 \times 25 \\ \implies n & = \boxed{23} \end{aligned}$