How many numbers satisfy b³ = b?

Challenge

b^{3} = b

Press play — the board writes itself.

0:00 / 1:33

Transcript

How many numbers satisfy b³ = b? Not which ones — how many. Most people find two. There are three… and the missing one is hiding in plain sight.

$b^{3} = b$

Here's the move everyone makes. A b on both sides? Divide both sides by b. You get b² = 1, so b is 1 or −1. Two answers — feels done, right?

$\div b : b^{2} = 1 \Rightarrow b = \pm 1 ?$

But dividing by b has fine print: it only works if b isn't zero. Now check zero in the original — zero cubed IS zero. It works! Dividing didn't simplify the equation… it deleted a solution.

So here's the safe move: never divide by something that could be zero. Bring everything to one side and factor — b, times b minus one, times b plus one, equals zero. A product is zero when any factor is zero… so each factor hands you an answer.

$b (b - 1) (b + 1) = 0 \Rightarrow b = - 1, 0, 1$

Look at them on the number line. Negative one… one… and zero, sitting right in the middle — the one dividing threw away.

So when a variable shows up on both sides, don't divide it away. Move everything to zero and factor. Factoring keeps every root — dividing can quietly delete one.

The solution dividing deletes · 1:33

The instinct is to divide both sides by $b$ — and that instinct quietly costs you a solution. Divide, and $b^3 = b$ collapses to $b^2 = 1$ , giving $b = \pm 1$ : two answers. But dividing by $b$ only makes sense when $b \ne 0$ , and $b = 0$ solves the original just fine, since $0^3 = 0$ .

Keep every solution by moving everything to one side instead of dividing, then factoring:

$b^3 - b = 0 \;\Rightarrow\; b(b^2 - 1) = 0 \;\Rightarrow\; b(b-1)(b+1) = 0$

A product equals zero exactly when one of its factors is zero, so $b = 0$ , $b = 1$ , or $b = -1$ — three solutions, sitting neatly at the center and one step to either side.

What this hides

Dividing both sides by a variable feels like fair play — it looks like “undoing” a multiplication, the same balancing move that solves an ordinary equation. The catch is that the move is only legal when the divisor isn’t zero, and $b = 0$ is exactly the root it throws away. You didn’t simplify the equation; you deleted part of it, and the deletion is invisible because the answer that survives still looks reasonable.

The transferable idea: to solve, move everything to zero and factor — and never divide by something that could be zero. Factoring keeps every root because it discards nothing. A product being zero is the whole engine, and each factor hands you one answer.