We know from Question 7.4.9(b) that \[(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3,\] but before mathematicians developed algebra, they came up with geometric explanations for facts like this. For example, the Renaissance mathematician Cardano would have visualised $(a+b)^3$ as the volume of a cube with side length $a+b$. The cube can be cut up along the lines dividing each edge into $a$ and $b$:

How many pieces will the cube be cut into? You might like to try to draw the pieces. Show answer

What is the volume of each of the pieces? Hence show that $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Show answer

The small cube has volume $a^3$, the long pieces each have volume $a^2b$, the flat pieces each have volume $ab^2$, and the large cube has volume $b^3$. So the total volume is $a^3 + 3a^2b + 3ab^2 + b^3$. This must be equal to the volume of the original cube of side length $a+b$, which was $(a+b)^3$.