In short

When you see (a+b+c)^2 on an exam paper, do not try to recall a six-term formula. Picture a 3 \times 3 grid. The 3 cells on the main diagonal give a^2 + b^2 + c^2. The 6 off-diagonal cells pair up into matched twins, contributing 2ab + 2bc + 2ca. Total: a^2 + b^2 + c^2 + 2(ab + bc + ca). Three squares plus three doubled cross-terms — that is the whole identity, recovered from a single mental picture in under five seconds.

You have seen the identity before. The full geometric dissection — slice a square of side a+b+c twice horizontally and twice vertically, count nine pieces — is the proper proof. What this article gives you is the exam-hall version: a sketch you can draw on the side of your paper in three seconds, or just hold in your head, that lets you write down the six terms without thinking.

This is the standard expansion CBSE Class 9 Algebra expects you to know cold. It also reappears all the way through JEE Main and Advanced — every time a problem hands you a+b+c and ab+bc+ca and asks for a^2+b^2+c^2, you are meant to invoke this identity in one line.

The mental picture

Draw a 3 \times 3 grid. Label the rows a, b, c from top to bottom. Label the columns a, b, c from left to right. Each cell holds the product of its row label and its column label.

A small 3 by 3 grid showing the mental shortcut for expanding (a + b + c) squaredA compact 3 by 3 grid with rows labelled a, b, c and columns labelled a, b, c. The three diagonal cells contain a squared, b squared, c squared, all shaded the same accent colour to mark them as the diagonal squares. The six off-diagonal cells contain ab, ac, ab, bc, ac, bc, paired by colour to show that each pair contributes a doubled cross-term. a b c a b c a² ab ac ab b² bc ac bc c²
The $3 \times 3$ grid you keep in your head. Diagonal cells (one accent colour) are the squares $a^2, b^2, c^2$. Each off-diagonal pair shares a colour and contributes a $2ab$, $2bc$, or $2ac$. Sketch this in the margin, count, write the answer.

That is the entire shortcut. Three diagonal squares, three pairs of matched off-diagonal cells. Read the answer straight off the picture:

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

Why do off-diagonal cells come in pairs? The cell in row a column b holds the product a \cdot b. The cell in row b column a holds b \cdot a. These are the same number — multiplication is commutative — but they are two different cells in the grid. So when you collect terms, ab shows up twice. Same for bc and ac. That is where every "2" in the cross-terms comes from.

Three quick worked examples

Example 1 — pure number check: $(2+3+4)^2$

You can compute this directly: 2 + 3 + 4 = 9, so (2+3+4)^2 = 81. Now use the grid to verify the identity is doing what it says.

  • Diagonal squares: 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29.
  • ab pair: 2 \times 3 = 6, doubled is 12.
  • bc pair: 3 \times 4 = 12, doubled is 24.
  • ac pair: 2 \times 4 = 8, doubled is 16.

Add: 29 + 12 + 24 + 16 = 81. Why this matters: notice that the three diagonal squares only gave you 29 out of 81. The cross-terms contributed 52 — more than half the answer. Forgetting them is the single biggest source of marks lost on this identity.

If a friend tells you (2+3+4)^2 = 4 + 9 + 16 = 29, you now have an instant counterexample to throw at them.

Example 2 — the symbolic case you will use most: $(x + y + z)^2$

Same grid, just relabelled. Diagonal cells: x^2, y^2, z^2. Off-diagonal pairs: xy twice, yz twice, zx twice. Read off the answer:

(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx

This is the form CBSE Class 9 textbooks ask you to memorise. With the grid in your head, you do not memorise it — you regenerate it in a heartbeat. Why this version is so common: in algebraic manipulations, x, y, z are the standard placeholder names, and this expansion is the workhorse identity behind every "find x^2 + y^2 + z^2 given x+y+z and xy+yz+zx" problem.

Example 3 — the classic exam question: find $a^2 + b^2 + c^2$

Given: a + b + c = 5 and ab + bc + ca = 6. Find: a^2 + b^2 + c^2.

Apply the identity with a, b, c in their standard slots:

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

Substitute the known values: 5^2 = (a^2 + b^2 + c^2) + 2 \times 6, so

25 = (a^2 + b^2 + c^2) + 12 \quad \Longrightarrow \quad a^2 + b^2 + c^2 = 13

Why this trick works without ever finding a, b, c individually: the identity links three symmetric quantities — the sum, the sum-of-pairwise-products, and the sum-of-squares — by a single equation. Knowing any two of them pins down the third, regardless of what the individual values are. (In fact, a, b, c here are the three roots of a cubic with sum 5 and pairwise product 6 — and you never needed to solve that cubic.)

This pattern is everywhere in JEE-level algebra. The moment you see two of {sum, sum-of-products, sum-of-squares} given, write down (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) and substitute. The answer drops out in one line.

Why the grid view scales beyond three terms

The picture is not special to 3 variables. For (a + b + c + d)^2, draw a 4 \times 4 grid — 16 cells. The 4 diagonal cells give a^2 + b^2 + c^2 + d^2. The remaining 12 off-diagonal cells pair up into 6 pairs, contributing 2(ab + ac + ad + bc + bd + cd). Why the count works: there are \binom{n}{2} unordered pairs of distinct variables among n, and each pair sits in two mirror-image cells across the diagonal. So you always get n squares plus \binom{n}{2} doubled cross-terms, no matter how many variables you have.

For n = 5 — say (a+b+c+d+e)^2 — picture a 5 \times 5 grid: 5 diagonal squares plus \binom{5}{2} = 10 doubled cross-terms. That is 15 distinct contributions, 25 cells in total. The general formula

\left(\sum_{i=1}^{n} a_i\right)^2 = \sum_{i=1}^{n} a_i^2 + 2 \sum_{i < j} a_i a_j

is the same grid argument written compactly.

How to use this in the exam hall

  1. See (\,\text{sum of 3 things}\,)^2. Pause for one second.
  2. Sketch a tiny 3 \times 3 grid in the margin — three rows and three columns, no labels needed if you already know which variables are involved. Or just hold it in your head.
  3. Write three diagonal squares: a^2 + b^2 + c^2.
  4. Add the three doubled cross-terms in cyclic order: 2ab + 2bc + 2ca.
  5. Done. The whole expansion is on your paper in under ten seconds.

If the question is a "find a^2 + b^2 + c^2 given the sum and the sum-of-products" problem, swap step 4 for the rearranged form

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

and substitute. Why this rearrangement is worth memorising separately: it is the form that appears most often in exam questions. Both forms are the same identity — one moves the cross-terms to the right side, the other to the left.

Common slips to watch for

References

  1. NCERT Class 9 Mathematics, Chapter 2: Polynomials — the CBSE textbook section that introduces (x+y+z)^2 as a standard identity.
  2. Wikipedia: Square of a sum — multi-variable case with the same grid intuition.
  3. Wikipedia: Multinomial theorem — generalises (a_1 + \cdots + a_n)^k to any n and k.
  4. Art of Problem Solving: Symmetric sums — how the sum, sum-of-products, sum-of-squares triangle of identities powers JEE algebra.