Why is $\,c^2-2bcd+b^2d^2=(c-bd)^2\,$?

Question

How would you explain, using simple arithmetic, that $$c^2-2bcd+b^2d^2=(c-bd)^2\;?$$ (I'm trying to explain this to a student I tutor.)

Answer 1

$$\begin{align} (c-bd)^2 &= (c-bd)(c-bd)\\ \\ & = c(c-bd) - bd(c-bd) \\ \\ & = c^2 -c(bd) - (bd)c + (bd)^2 \\ \\ & = c^2 - 2bcd +b^2d^2\end{align}$$

I've simply used the distributive property of multiplication over addition, the associative and commutative properties of multiplication on the reals, and the fact that $(bd)^2 = b^2d^2$.