Identities in algebra can sometimes be shown as area relationships. For example:

The figure shown corresponds to the identity $(a + b)^2 = a^2 + 2ab + b^2$. Do you see how?
Draw figures corresponding to the identities $(a + b)(a - b) = a^2 - b^2$ and
  $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

Identity 1: $(a + b)(a - b) = a^2 - b^2$

We need two rectangles: one of size $(a + b) \times a$ and one of size $(a + b) \times b$, then subtract the second from the first.

Step 1: Draw a rectangle of width $(a + b)$ and height $a$:

$$\text{Area} = a \times a + b \times a = a^2 + ab$$

Step 2: Draw a rectangle of width $(a \times b)$ and height $b$ and subtract:

$$\text{Area} = a \times b + b^2 = ab + b^2$$

Combined area model $(a + b) \times (a - b)$:

Blue $(a^2)$ = kept area.

Red $(ab, ab)$ = two equal rectangles that cancel.

Orange $(b^2)$ = removed.

$$\text{Net Area} = a^2 + ab - ab - b^2 = a^2 - b^2$$

$$(a + b)(a - b) = a^2 - b^2$$

Identity 2:

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

Represent $(a + b + c)^2$ as the area of a square with side $(a + b + c)$. Divide each side into three parts $a, b, c$. This creates a $3 \times 3$ grid of nine rectangles:

Blue squares $(a^2, b^2, c^2)$ lie on the diagonal.

Green $(ab, ab) \rightarrow 2ab$.

Orange $(bc, bc) \rightarrow 2bc$. Yellow $(ca, ca) \rightarrow 2ca$.

$$\text{Total Area} = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$