2013 AMC 10B Problem 14

1. Write out both sides using the definition $a \clubsuit b = a^2 b - a b^2$:
$$x \clubsuit y = x^2 y - x y^2, \qquad y \clubsuit x = y^2 x - y x^2.$$
2. Set them equal and move everything to the left side:
$$x^2 y - x y^2 - y^2 x + y x^2 = 0, \quad\text{that is,}\quad 2x^2 y - 2x y^2 = 0.$$
3. Every term contains $2xy$, so pull it out:
$$2xy(x - y) = 0.$$
4. A product is $0$ exactly when at least one factor is $0$. So a point works when
- $x = 0$ (the $y$-axis), or
- $y = 0$ (the $x$-axis), or
- $x = y$ (the slanted line through the origin).
5. Test a few points. $(0, 7)$: both sides are $0$. $(4, 0)$: both sides are $0$. $(3, 3)$: both sides are $27 - 27 = 0$. But $(1, 2)$: $1 \clubsuit 2 = 2 - 4 = -2$ while $2 \clubsuit 1 = 4 - 2 = 2$, so a point off the three lines does not work.
6. These are three different lines.

The answer is $\boxed{\textbf{(E)}\ \text{three lines}}$.

A made-up operation is just an ordinary expression in disguise, and factoring shows what it really says. When a product equals $0$, each factor gives its own set of solutions. Never divide by an expression like $xy$ that could be zero: each factor you cancel is a whole line of answers thrown away.

First factor the operation: $a \clubsuit b = ab(a - b)$. Then
$$x \clubsuit y = xy(x - y), \qquad y \clubsuit x = yx(y - x) = -xy(x - y).$$
Setting them equal gives $xy(x - y) = -xy(x - y)$, i.e.
$$2xy(x - y) = 0.$$
A product is zero when some factor is zero, so the solution set is the union of $x = 0$ (the $y$-axis), $y = 0$ (the $x$-axis), and $x = y$. These are three distinct lines, all passing through the origin.

The answer is $\boxed{\textbf{(E)}\ \text{three lines}}$.