In short

(a+b+c)^2 is the area of a square whose side is a+b+c. Slice that square twice horizontally and twice vertically — at the points where a ends and where b ends — and the big square breaks into a 3 \times 3 grid of nine rectangles. The three pieces on the main diagonal are squares of areas a^2, b^2, c^2. The remaining six off-diagonal pieces pair up into matching twins of total area 2ab, 2bc, and 2ca. The identity (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca is just the equation "big square area = sum of nine piece areas". Why: every off-diagonal piece appears twice — once in row i column j, once in row j column i — because area is symmetric in its two side-lengths.

You already know the two-variable case: (a+b)^2 slices a square into four pieces. Add a third variable c and the same trick scales up — but now you need two horizontal cuts and two vertical cuts, and the square breaks into nine pieces instead of four. The number of a^2-style "diagonal" pieces is three (one for each variable), and the number of off-diagonal "rectangle" pieces is six, which pair up into the three cross-terms 2ab, 2bc, 2ca.

This is the picture behind the formula every mensuration problem uses when an area or volume is written as a sum of three lengths squared, and the picture behind the JEE-style trick of "completing the square in three variables" that turns a quadratic form into a sum of squares.

The picture

Draw a square. Mark off lengths a, b, c along the top edge — so the side of the square is a+b+c. Do the same down the left edge. Now extend each mark across the square. You get two horizontal lines and two vertical lines, slicing the big square into a 3 \times 3 grid of nine rectangles.

Number the rows and columns 1, 2, 3, where row 1 has height a, row 2 has height b, row 3 has height c — and the same for columns. The piece in row i, column j has area equal to row-i height times column-j width. Three of those pieces sit on the main diagonal (1,1), (2,2), (3,3) and they are squares: a \times a, b \times b, c \times c. The other six are rectangles, and they come in matching pairs.

Geometric proof of (a + b + c) squared as a 3 by 3 grid of nine piecesA large square of side a plus b plus c, sliced into a 3 by 3 grid of nine rectangles. The three pieces on the main diagonal are squares of areas a squared, b squared, c squared, all shaded the same wheat colour. The two ab pieces (row a column b, and row b column a) are shaded matching light blue. The two bc pieces are shaded matching pale green. The two ca pieces are shaded matching light yellow. a² ab ca ab b² bc ca bc c² a b c a b c
The big square has side $a+b+c$, so its area is $(a+b+c)^2$. The nine pieces inside have areas $a^2, ab, ca, ab, b^2, bc, ca, bc, c^2$. The three diagonal pieces are squares; the six off-diagonal pieces match in three coloured pairs. Total: $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

That is the entire proof. The diagonal squares give you a^2 + b^2 + c^2. Each pair of matching-colour rectangles gives you a 2ab, 2bc, or 2ca — twice the rectangle area, because the pair has two copies of it. Add it all up and you have the identity.

Why does every off-diagonal rectangle come in a pair? Because rectangle area is symmetric: a rectangle of width a and height b has the same area as a rectangle of width b and height a. In the grid, "row a, column b" sits in the top-right region, and "row b, column a" sits in the middle-left. Two physical pieces, same area each, so together they contribute 2ab.

Worked examples

Take a = 3, b = 4, c = 5. Then a + b + c = 12, so the big square has side 12 and area 144. Now count the nine pieces:

  • diagonal squares: a^2 = 9, b^2 = 16, c^2 = 25. Sum: 50.
  • ab pair: each piece is 3 \times 4 = 12, two of them give 24.
  • bc pair: each piece is 4 \times 5 = 20, two of them give 40.
  • ca pair: each piece is 5 \times 3 = 15, two of them give 30.

Add: 50 + 24 + 40 + 30 = 144. The picture and the arithmetic agree. And 144 = 12^2, which is what (3+4+5)^2 should be.

If you had naively written (3+4+5)^2 = 9 + 16 + 25 = 50, you would have missed six whole rectangles — 94 out of 144 in this example, almost two-thirds of the square. The cross-terms are not a small correction; in three variables they often dominate.

Why every cross-term comes with a 2. Look at the grid again. The piece in row a, column b is a rectangle of width b and height a, area ab. The piece in row b, column a is a rectangle of width a and height b, also area ab. These are two different physical pieces of the big square — one sits in the upper-right region, the other in the middle-left region. They have the same area but they are not the same rectangle.

So when you collect terms, the ab piece shows up twice in the sum, giving 2ab. The same happens for bc (one piece in row b column c, another in row c column b) and for ca (row c column a, and row a column c). Why: in a grid where rows and columns are both labelled by the same set of variables, every off-diagonal cell (i, j) has a mirror cell (j, i) across the main diagonal. There are exactly \binom{3}{2} = 3 unordered pairs of distinct variables — \{a,b\}, \{b,c\}, \{c,a\} — and each pair contributes two matching rectangles to the grid.

This is also why algebraists say the cross-terms in (a+b+c)^2 live "off the diagonal": they correspond exactly to the off-diagonal cells of a 3 \times 3 multiplication table whose rows and columns are a, b, c.

Extending to four variables: (a+b+c+d)^2. The same picture scales up. With four variables you slice the big square three times horizontally and three times vertically, producing a 4 \times 4 grid of 16 pieces. Of those, 4 sit on the diagonal and are squares: a^2, b^2, c^2, d^2. The other 16 - 4 = 12 are off-diagonal rectangles, and they pair up into 12 / 2 = 6 pairs, giving the cross-terms 2ab, 2ac, 2ad, 2bc, 2bd, 2cd.

So (a+b+c+d)^2 = a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd. Four squares plus six doubled cross-terms — eleven distinct contributions, 16 pieces total.

The number of cross-term pairs in the n-variable case is \binom{n}{2} = \dfrac{n(n-1)}{2}, because that is the number of ways to pick two distinct variables out of n to form an unordered pair. For n = 2 you get 1 pair (the lone 2ab in the standard identity), for n = 3 you get 3 pairs, for n = 4 you get 6, and so on. The general identity is

(a_1 + a_2 + \cdots + a_n)^2 = \sum_{i=1}^{n} a_i^2 + 2 \sum_{1 \le i < j \le n} a_i a_j

— the n diagonal squares and the \binom{n}{2} doubled cross-terms.

Where this identity shows up

Mensuration problems. Whenever the perimeter of a rectangle, or the dimensions of a box, are written as a + b + c (sometimes hidden as l + b + h or x + y + z), squaring that sum invokes this identity. A box of dimensions l, b, h has surface area 2(lb + bh + hl) — exactly the cross-term part of (l+b+h)^2 - (l^2 + b^2 + h^2). So if you know the perimeter-like quantity l+b+h and the diagonal \sqrt{l^2 + b^2 + h^2}, the identity instantly gives you the surface area without solving for the individual sides.

Completing the square in three variables. A quadratic form like x^2 + y^2 + z^2 + 2xy + 2yz + 2zx is exactly (x+y+z)^2. Spotting this collapses a six-term mess into a single squared bracket. JEE Advanced problems often hide this pattern with shifted variables: x^2 + y^2 + z^2 - 2xy + 2yz - 2zx is (x - y - z)^2 if you are willing to track the signs — and the same nine-piece picture, with some pieces signed negative, justifies it.

Identity for a^2 + b^2 + c^2. Rearranging the main identity gives

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

a workhorse formula in algebra problems where you know the sum and the sum-of-products of three numbers (often the roots of a cubic) and need their sum-of-squares. The dissection picture is the geometric reason this rearrangement is true — you are subtracting the six off-diagonal rectangles from the whole square to isolate the three diagonal squares.

Going deeper

The 3 \times 3 grid is the case n = 3 of a much bigger object — the multiplication table of the polynomial expansion (a_1 + a_2 + \cdots + a_n)(a_1 + a_2 + \cdots + a_n), where every term in the first bracket is multiplied by every term in the second. The result is an n \times n table of n^2 products a_i a_j, and collecting like terms gives the diagonal a_i^2 contributions and the off-diagonal 2 a_i a_j contributions.

In linear algebra you eventually learn to write this as \mathbf{v}^T \mathbf{v} where \mathbf{v} = (a_1, a_2, \ldots, a_n), and the matrix of all products a_i a_j is called the outer product \mathbf{v}\mathbf{v}^T. The diagonal of that matrix gives the squared terms and the symmetric off-diagonal entries give the cross-terms — exactly the structure of the dissection picture, generalised to n dimensions.

The same nine-piece dissection also extends to cubes. The identity (a+b+c)^3 slices a cube of side a+b+c into a 3 \times 3 \times 3 grid of 27 rectangular boxes, of which 3 are perfect cubes (a^3, b^3, c^3) on the main space-diagonal, 18 are "rectangular boxes with two short sides equal" that group into 6 matching triples giving 3a^2 b, 3ab^2, 3b^2 c, 3bc^2, 3a^2 c, 3ac^2, and 6 are "rectangular boxes with all three side-lengths different" that group into one matching sextuple giving 6abc. That is where the famous 6abc cross-term in (a+b+c)^3 comes from — it is six identical a \times b \times c boxes scattered through the cube.

References

  1. Wikipedia: Square of a sum — covers the two-variable and multi-variable cases of the same identity.
  2. Wikipedia: Multinomial theorem — the generalisation of (a+b+c)^n for arbitrary n.
  3. NCERT Class 9 Mathematics, Chapter 2: Polynomials — the Indian school textbook section that introduces (x+y+z)^2 with a near-identical area diagram.
  4. Cut the Knot: Visual proofs of algebraic identities — a small gallery of dissection proofs including the three-variable case.
  5. Art of Problem Solving: Symmetric sums — how the cross-term pattern in (a+b+c)^2 generalises to higher symmetric polynomials.