(a + b + c)² by substitution, and the square that proves it

What to assume they know

• (a + b)² = a² + 2ab + b², proved rather than remembered (An identity holds for every value; an equation need not)
• Reading an identity off a partitioned square (Reading (a + b)² off a partitioned square)
• a² − b² = (a + b)(a − b) from Class 8, which the chapter recalls rather than re-derives
• Substituting one expression for a letter, and substituting back
• Area of a rectangle, and that cutting a region up and moving the pieces does not change total area

What they should be able to do

• Obtain (a + b + c)² by replacing b + c with a single letter and applying the two-term identity
• Unwind the substitution and collect the six terms of the expansion
• Write the result in the order that makes it memorable — the three squares first, then the three doubled products
• Label the nine pieces of a supplied 3 × 3 partitioned square so that the drawing represents the identity
• Explain from the figure why each pair of letters contributes twice while each single letter contributes once
• Square a three-digit number by splitting it into hundreds, tens and units
• Choose, for a given number, which of the three identities makes the squaring easiest, and say why
• Rewrite a² − b² = (a + b)(a − b) as a rule for computing a², and use it on a number ending in 5
• Justify that rewriting by cutting a square, turning one strip and reassembling
• Decide whether a proposed three-square statement is an identity, and correct it if it is not

Where it usually goes wrong

• "A new identity needs a new proof." It needs a new proof only if you cannot reduce it to an old one. The substitution d = b + c is the cheapest proof in the chapter and it assumes nothing beyond the two-letter identity.
• "(a + b + c)² = a² + b² + c²." The same error as the two-letter case, three times over. The figure makes the missing pieces visible: six of the nine cells are not squares at all.
• "There should be three cross terms, one per letter." There are three, but one per pair, not one per letter — ab, bc, ca. With four letters there would be six. Count the off-diagonal cells and the rule becomes obvious.
• "2ca and 2ac are different terms." They are the same term; the page writes the second form because it reads round the triangle a → b → c → a. Say so once, or students will look for a seventh term.
• "With a minus in the middle, (a − b + c)² needs its own identity." It does not. Take b as negative and the same six-term expansion applies, which is what makes Q2 (v) and Q3 (ii) tractable.
• "Q4 must be an identity, because it is printed in the book." It is printed as a question. One substitution — a = b = c = 1 — refutes it. This is where the chapter's earlier lesson about a single counter-case pays off.
• "The three-term identity is always the best tool for a three-digit square." For 117 the two-term minus identity is faster. Exercise Set 4.3 Q1 is explicitly a question about which tool to reach for.
• "a² = (a + b)(a − b) + b² is a different fact from a² − b² = (a + b)(a − b)." It is the same fact with one term moved. What changes is what it is for: the first form computes squares, the second factorises differences.
• "Fig. 4.5 shows areas being lost." Nothing is lost. One strip is turned through a right angle and re-laid; the b-by-b corner is the only piece that stays where it was, and it is the whole reason for the +b².

Questions to check understanding

• Expand a three-term square, including cases where one or two of the terms are negative
• Factorise a six-term expression by recognising it as a three-term square
• Square a three-digit number, stating which identity was chosen and why
• Label the pieces of a supplied 3 × 3 partitioned square to represent the identity
• Decide whether a proposed equality is an identity, and if not, give one set of values that breaks it
• Compute a square of a number ending in 5 by the (a + b)(a − b) + b² route, and generalise the pattern
• Justify a stated rearrangement identity from a supplied dissection figure

Examples worth working on the board

Inputs, not answers. Values marked Verified are worked out here; this book prints no answer key.

• The substitution (§4.4, p. 75). The page sets d = b + c, invokes (a + d)² = a² + 2ad + d², and then puts b + c back in place of d, giving a² + 2a(b + c) + (b + c)². Verified: the result is a² + 2ab + 2ac + b² + 2bc + c², which the page then reorders as a² + b² + c² + 2ab + 2bc + 2ca and recommends remembering in that order.
• Two cartoon panels carry the idea (p. 75, artwork). A drawn girl's thought bubble proposes putting d = b + c, and a drawn boy's bubble asks whether (a + b + c)² is (a + d)². Both are hand lettering inside artwork and neither extracts. They are worth keeping in the explanation's structure — the substitution is presented as somebody's idea rather than as a rule handed down — but the lettering should be rebuilt as type.
• Fig. 4.4 (§4.4, p. 76). A square partitioned by two horizontal and two vertical cuts into a 3 × 3 arrangement of nine pieces. The top edge is marked a, b, c and the left edge is marked a, b, c. The nine interior pieces carry no labels at all; the Think and Reflect box directly beneath asks the reader to supply them. What the figure does carry is a colour code: the three pieces on the diagonal have three distinct fills, and each of the three off-diagonal pairs shares a fill with its partner. So the drawing has already told you which pieces are equal before you have written anything. Verified as a statement: the diagonal pieces are a², b², c²; the two pieces in the a-row/b-column and b-row/a-column are both ab; likewise ac and bc; the nine areas sum to (a + b + c)².
• Example 9 (§4.4, p. 76). Square 119 by splitting it as 100 + 10 + 9. Inputs: the three parts. Verified: 10000 + 100 + 81 + 2000 + 1800 + 180 = 14161. The page prints the six pieces before adding them, which is the right way round for showing it.
• The roll-up (§4.4, p. 76). The page numbers the three identities verified so far — the sum of two squared, the difference of two squared, and the sum of three squared — and says they have been used both for calculation and for manipulating expressions. This list is the spine of Exercise Set 4.3.
• Exercise Set 4.3 Q1 (p. 76), six squares to compute, with the question asking which identity makes each easiest: 117², 78², 198², 214², 1104², 1120². Verified: 78² = (80 − 2)² = 6400 − 320 + 4 = 6084 and 198² = (200 − 2)² = 40000 − 800 + 4 = 39204, both cleanest with the minus identity; 1104² = (1100 + 4)² = 1210000 + 8800 + 16 = 1218816 and 1120² = (1100 + 20)² = 1210000 + 44000 + 400 = 1254400, both cleanest with the plus identity; 214² = (200 + 10 + 4)² = 40000 + 100 + 16 + 4000 + 1600 + 80 = 45796, which wants the three-term identity; and 117² is the interesting one, because (100 + 10 + 7)² gives 10000 + 100 + 49 + 2000 + 1400 + 140 = 13689 while (120 − 3)² gives 14400 − 720 + 9 = 13689 with three terms instead of six. The exercise's real question is a judgement about effort, not about correctness.
• Exercise Set 4.3 Q2 and Q3 (p. 77). Q2's three-letter items: m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n² and 9a² + 4b² + c² − 12ab + 6ac − 4bc. Q3 asks for two expansions: (p + 3q + 7r)² and (3x − 2y + 4z)². Verified: the first Q2 item is (m/3 + k/2 + 3n)²; the second is (3a − 2b + c)², where the signs work out because exactly one of the three letters, b, carries the minus — what comes in twos is the negative cross terms, −12ab and −4bc, each of which contains that b, while 2ca stays positive since neither of its letters is the negative one; (p + 3q + 7r)² = p² + 9q² + 49r² + 6pq + 42qr + 14rp; and (3x − 2y + 4z)² = 9x² + 4y² + 16z² − 12xy − 16yz + 24zx.
• Exercise Set 4.3 Q4 (p. 77). The question asks whether (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c². Verified: it is not an identity. Expanding, the left side comes to 3a² + 3b² + 3c² − 2ab − 2ac − 2bc, and the cross terms do not vanish. Try a = b = c = 1: the left side is 1 + 1 + 1 = 3 and the right side is 6. One counter-case settles it, which is the discipline established in An identity holds for every value; an equation need not.
• a² rewritten (§4.4, p. 77). The page recalls a² − b² = (a + b)(a − b) from Class 8 and rearranges it as a² = (a + b)(a − b) + b², observing that in this form it computes a square. Inputs only; the rearrangement is one step.
• Fig. 4.5 (§4.4, p. 77). Read off the printed page, because none of the lettering extracts. The figure shows a square of side a, marked a across the top and a down the left. A strip of height b along the bottom is drawn dotted; its left portion, of width a − b, is shown being carried by a curved arrow round to the right-hand side of the figure, where it reappears as a vertical strip of width b and height a − b in a contrasting fill. What is left standing is a rectangle of width a + b and height a − b, plus the b-by-b corner square that was never moved. Verified as a statement: moving a strip does not change area, so a² = (a + b)(a − b) + b², which is exactly the rearranged identity the caption asks the reader to justify. The b² corner is the piece that is left over, and it is why the identity has a +b² on the end.
• Śhrīdharāchārya's method (§4.4, p. 77, side box). Dated 750 CE in the book. Worked example printed: 55² computed as (55 + 5)(55 − 5) + 5², i.e. 60 × 50 + 25. Verified: 3000 + 25 = 3025. The choice b = 5 is what makes it work — it pushes one factor to a multiple of 10 and pulls the other down to a multiple of 10 at the same time.
• Think and Reflect, four squares and a pattern (p. 78). The box asks for 35², 65², 85², 105² by a suitable identity and then asks whether anything interesting shows up. Verified by Śhrīdharāchārya's form with b = 5: 35² = 40 × 30 + 25 = 1225; 65² = 70 × 60 + 25 = 4225; 85² = 90 × 80 + 25 = 7225; 105² = 110 × 100 + 25 = 11025. The pattern, which the book does not state: a number ending in 5 is 10n + 5, and its square is n(n + 1) hundreds plus 25 — 3 × 4 = 12 gives 1225, 6 × 7 = 42 gives 4225, 8 × 9 = 72 gives 7225, 10 × 11 = 110 gives 11025. This is the strongest single beat in the section and it is unprinted, so the explanation owns it.
• Fig. 4.6 (p. 78, inside the same Think and Reflect box). Two rows of squares, read off the printed page. The top row has four squares, whose sides are labelled a + b + c, a + b − c, a − b + c and a − b − c, in that order, and the leftmost of them is itself cut into pieces — but not into a tiling, and this is worth getting right before anyone redraws it. It holds a square of side 2b in one corner, a square of side 2c in the opposite corner, and two congruent rectangles of (a + b − c) by (a − b + c); those two rectangles cross each other, sharing a small square of side a − b − c. Nothing fills a remaining gap: the two corner squares do not even reach across the side, since 2b + 2c falls short by exactly the overlap's side. So the top-left square is accounted for by a subtraction, not by addition alone: take the two corner squares, (2b)² and (2c)²; add twice the area of the crossing rectangle, whose sides measure (a + b − c) and (a − b + c); then take away, once, the shared square whose side measures a − b − c. That total is the area of the whole, and the piece subtracted is precisely the fourth square of the top row — which is what makes the two rows balance. Measured on the printed page, then re-derived: at the drawn proportions the side reads 485 px, 2b = 272, 2c = 150, so a = 274 and the overlap square is 63 px, and the rectangles come out 335 by 213 against a measured 340 by 215. The bottom row has three squares, labelled 2a, 2b and 2c; the 2a square is cut into four quadrants by one horizontal and one vertical line, and a small square sits at the junction. Colours match across the two rows, piece for piece. The box says only that the figures represent an identity and invites the reader to identify it; no identity and no answer is printed anywhere in pp. 68–91. Verified by an added expansion: the identity is (a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = (2a)² + (2b)² + (2c)². Every cross term cancels in fours, leaving 4a² + 4b² + 4c² on both sides. The figure's dissection is the same statement done with scissors: the two red rectangles are common to both rows and cancel, the 2b and 2c squares sit on one side, and the three remaining squares — a + b − c, a − b + c and a − b − c — sit on the other. Pair this with Q4: the three-square statement Q4 offers is false, and Fig. 4.6 shows what the true four-square statement is.

Figures to have open

• Fig. 4.4, the 3 × 3 partitioned square (p. 76). The chapter's own figure and indispensable. Redraw as a schematic. Keep the colour pairing — it is the figure's argument — and keep the interior blank at first, since the book's Think and Reflect is precisely to fill it in.
• Fig. 4.5, the cut-and-turn rearrangement (p. 77). The chapter's own figure. It is worth showing as a movement rather than as a still; as a still image it is hard to read, and the arrow indicating the moved strip is the only clue that anything moves.
• Fig. 4.6, the two rows of squares (p. 78). The chapter's own figure and the hardest thing on the page. Redraw at generous size, keep the colour matching between the rows, and show the cancellation of the two shared rectangles; without that step the figure is a decoration.
• A column table for the numbers-ending-in-5 pattern (35, 65, 85, 105 against n(n + 1) and the constant 25). Standard schematic, not in the book.
• The two cartoon panels on p. 75 need not be reproduced; rebuild their two lines as clean type.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 4, "Exploring Algebraic Identities": §4.4, printed as "4.4. More Identities", pp. 75–78, including Example 9, Fig. 4.4, Fig. 4.5 and Fig. 4.6.
• Exercise Set 4.3, pp. 76–77, all four questions.
• Think and Reflect boxes: p. 76, which asks for Fig. 4.4 to be labelled; p. 78, which asks for four squares of numbers ending in 5 and then poses Fig. 4.6.
• The Śhrīdharāchārya side box, p. 77, dated 750 CE, with 55² worked.
• Chapter summary, p. 90; the identity list on p. 91 includes both the three-term square and (x + y)(x − y) = x² − y².
• Companion topics: An identity holds for every value; an equation need not for the counter-case discipline Q4 needs, Reading (a + b)² off a partitioned square for the two-letter square this section extends.