Cubes: (a ± b)³ from a cube cut into eight pieces

What to assume they know

• (a + b)² = a² + 2ab + b², derived rather than recalled (An identity holds for every value; an equation need not)
• Multiplying a three-term expression by a two-term expression using distributivity
• Volume of a cube as edge cubed, and of a cuboid as the product of three dimensions (Predicting What Comes Next: Exploring Sequences and Progressions treats solids; the chapter assumes the formulae)
• That (−b)² is positive and (−b)³ is negative (Why a debt times a debt is a fortune)
• Reading a three-dimensional drawing of a solid cut into blocks

What they should be able to do

• Multiply (a + b) by a² + 2ab + b² and collect the result into four terms
• Identify the eight pieces of a cube of edge a + b and state the volume of each
• Explain why there are exactly three pieces of volume a²b and three of volume ab², by counting which direction carries the b
• Add the eight volumes and read the identity off the solid
• Obtain (a − b)³ from (a + b)³ by replacing b with −b, term by term
• Explain why the four signs alternate rather than all four flipping
• Given a cubic expression, decide whether it is a cube of a binomial and if so find the binomial
• Recover the edge of a cube from an expression for its volume
• Cube a three-digit number by splitting it at a round value, including negative cases
• State the conditions under which the solid picture is drawable, and what the algebra covers beyond them

Where it usually goes wrong

• "(a + b)³ = a³ + b³." The volume argument destroys this on sight: the two corner cubes account for only part of the solid, and the six boxes between them are the whole reason the identity has four terms.
• "1, 3, 3, 1 is something you learn by heart." It is a count of choices. Show the eight pieces being generated by three yes/no decisions, and the numbers appear on their own.
• "There are three a²b boxes because the cube has three faces you can see." No — because there are three dimensions, any one of which can be the b. The drawing hides three of the six boxes behind the visible ones, which is exactly why the exploded panel of Fig. 4.10 exists.
• "Squares give 1, 2, 1 and cubes give 1, 3, 3, 1, so fourth powers give 1, 4, 4, 1." The direction-count argument gives 1, 4, 6, 4, 1. Do not invite the fourth power unless the explanation is willing to do that count properly.
• "Replacing b by −b flips all four signs." It flips the two terms with an odd power of b. 3a(−b)² is positive.
• "(a − b)³ = a³ − b³." Different identity entirely, and it belongs to Sum and difference of cubes, and the three-term cubic identity. Keep the two apart, because a student who conflates them will factorise a³ − b³ as (a − b)³.
• "The cube picture proves the identity for every a and b." It proves it for positive lengths with b smaller than a. The algebraic derivation on p. 82 — which is only distributivity — is what covers the rest, exactly as with the square in §4.2.
• "Fractions break the identity." Q1 (viii) and (ix) put fractions in the b slot and nothing changes. The derivation never assumed anything about a or b.

Questions to check understanding

• Expand a cube of a binomial where one or both parts carry a coefficient or a fraction
• Identify a four-term expression as a cube of a binomial and state the binomial
• Recover the edge of a cube from a given expression for its volume
• Cube a three-digit number by splitting it at a round value, including negative numbers
• Factorise a four-term cubic whose terms are printed out of order
• Explain why the expansion of (a + b)³ has coefficients 1, 3, 3, 1, using the cut cube

Examples worth working on the board

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

• James and Reshma (§4.6, p. 82, Think and Reflect). Two students expand (a − b)²(a + b) differently. James substitutes the expansion of (a − b)² and multiplies out. Reshma regroups the three brackets as (a − b) times [(a − b)(a + b)] and uses the difference of squares to get (a − b)(a² − b²). The box asks who is correct and why, and invites the reader to combine identities for new results. No answer is printed. Verified: both are correct and both give a³ − a²b − ab² + b³. Reshma's route is shorter, and the reason it is legal is associativity of multiplication — you may bracket three factors either way. This is the perfect opening beat for §4.7, because the section's whole method is combining identities you already have.
• The cube identity, derived (§4.7, p. 82). (a + b)³ is (a + b) times (a² + 2ab + b²). Inputs: those two brackets. Verified: a³ + 3a²b + 3ab² + b³. Note that the chapter states this derivation twice — once here on p. 82 and again on p. 84 after the figure — with the same working. A teacher should treat it as one derivation, framed either side of the picture.
• Fig. 4.9 (§4.7, p. 82). A small line drawing of a cube, with each of three visible edges marked a then b, so the edge is a + b in every direction. Read on the printed page; the labels are artwork lettering. This is the figure that poses the question rather than answering it.
• Fig. 4.10 (§4.7, p. 83), read off the printed page. Three panels. The first shows the cube of edge a + b with the a-by-a-by-a corner block shaded, so the a³ piece is identified in place. The second shows the same cube with the three a-by-a-by-b blocks in one fill and the three a-by-b-by-b blocks in another, with the a and b edge labels marked on each face. The third panel, running the full width, is the exploded view: the a³ cube, then a plus sign, then the three a²b blocks bracketed and labelled 3a²b, then a plus sign, then the three ab² blocks bracketed and labelled 3ab², then a plus sign, then the small b³ cube. Verified as a statement: eight pieces in all, two of them cubes and six of them cuboids, and the volumes sum to (a + b)³.
• The printed accounting (§4.7, p. 83). The page states that three of the six cuboids measure a by a by b and the other three measure a by b by b, which makes their volumes a²b and ab² respectively, and that the six together come to 3a²b + 3ab².
• The direction-count argument. Not printed; this is added here, and it is the topic's thesis. Each piece is fixed by deciding, independently for the length, the width and the height, whether that dimension is a or b. Three decisions with two options each give eight pieces. All three a gives a³ — one way. Exactly one b gives a²b — three ways, one for each dimension that could be the b. Exactly two b gives ab² — three ways. All three b gives b³ — one way. So 1, 3, 3, 1. The same argument on a square gives 1, 2, 1, which is why 2ab has a 2 and not a 3, and it is worth showing the two counts side by side.
• Replacing b by −b (§4.7, p. 84). The page writes [a + (−b)]³ and substitutes into the identity, giving a³ + 3a²(−b) + 3a(−b)² + (−b)³. Inputs: the four substituted terms. Verified: a³ − 3a²b + 3ab² − b³. The page then observes that two terms are positive and two negative and that they alternate. Mechanism: the term carrying bᵏ picks up a factor (−1)ᵏ, so the even powers of b survive unchanged and the odd powers flip. That is why it alternates rather than all four turning negative.
• Example 13 (§4.7, p. 84). Find the edge of a cube whose volume, in cubic units, is p³ + 6p²q + 12pq² + 8q³. Inputs: those four terms. Verified: rewriting as p³ + 3p²(2q) + 3p(2q)² + (2q)³ identifies a = p and b = 2q, so the edge is p + 2q. The middle two terms are the test, as always: 3p²(2q) = 6p²q and 3p(4q²) = 12pq².
• Example 14 (§4.7, p. 84). Write 8n³ − 60n²m + 150nm² − 125m³ in the form (a − b)³. Inputs: the four terms. Verified: a = 2n and b = 5m, since 3(2n)²(5m) = 60n²m and 3(2n)(5m)² = 150nm², so the expression is (2n − 5m)³. Point worth making: both end terms are cubes of something (8n³ and 125m³), and the middle terms then decide, exactly as in the two-term case.
• Cubing numbers (End-of-Chapter Q2, p. 89). Five cube items: 147³, 199³, 127³, (−107)³, (−299)³. Verified: 147³ = (150 − 3)³ = 3375000 − 202500 + 4050 − 27 = 3176523; 199³ = (200 − 1)³ = 8000000 − 120000 + 600 − 1 = 7880599; 127³ = (130 − 3)³ = 2197000 − 152100 + 3510 − 27 = 2048383; (−107)³ = −(100 + 7)³ = −(1000000 + 210000 + 14700 + 343) = −1225043; (−299)³ = −(300 − 1)³ = −(27000000 − 270000 + 900 − 1) = −26730899. The two negative items are only a sign pulled out front, since a cube of a negative is the negative of the cube — but the split still has to be chosen inside the bracket, so 299 wants 300 − 1 and 107 wants 100 + 7.
• Cubes to factorise (End-of-Chapter Q3, p. 89). Items (v) and (xi) are four-term cubes with fractional parts: 27u³ − 1/125 − 27u²/5 + 9u/25 and 27u³ − 1/216 − 9u²/2 + u/4, both printed with their terms out of order. Verified: the first is (3u − 1/5)³ and the second is (3u − 1/6)³. Reordering before matching is the first move, exactly as in the two-term case.
• Q1's cube items (End-of-Chapter, p. 88). (x − y/3)³ and ((7/2)k − (2/3)m)³. Inputs only. Both are direct substitutions into the minus identity, with fractions in the b slot, and they are worth doing because students expect the identity to fail once fractions appear.

Figures to have open

• Fig. 4.10, the cut cube in three panels (p. 83). The chapter's own figure and indispensable. It is worth showing as a movement: the assembled cube, the cuts, then the exploded grouping into a³, three a²b, three ab², b³. A static redraw is much weaker than the printed original, which already uses three panels to do the work of a movement.
• Fig. 4.9, the cube of edge a + b (p. 82). Trivial to redraw and needed as the question before the answer.
• A generation tree for section 5 — three binary choices, eight leaves, the leaves grouped by how many b they contain. Standard schematic, not in the book, and the topic's most valuable single image.
• A side-by-side of the square's 1, 2, 1 against the cube's 1, 3, 3, 1. Standard schematic.
• No photograph is needed. Physical blocks would work if you have them, but the exploded drawing is enough.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 4, "Exploring Algebraic Identities": §4.7 "Finding New Identities", pp. 82–84, including Fig. 4.9, Fig. 4.10, Example 13 and Example 14.
• Think and Reflect, p. 82 — James and Reshma, printed at the end of §4.6 and read here as the doorway into §4.7.
• End-of-Chapter Exercises: Q1 (viii), (ix), p. 88; Q2 (iv)–(viii), p. 89; Q3 (v), (xi), p. 89.
• Chapter summary, p. 90; the identity list on p. 91 carries both cube identities.
• Companion topic: Sum and difference of cubes, and the three-term cubic identity for the sum and difference of cubes, which are a different pair of identities and are routinely confused with these.