Sum and difference of cubes, and the three-term cubic identity

What to assume they know

• Multiplying a two-term expression by a three-term expression, term by term
• (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx ((a + b + c)² by substitution, and the square that proves it)
• a² − b² = (a + b)(a − b), and the difference between it and (a − b)²
• The cube identities (a ± b)³ (Cubes: (a ± b)³ from a cube cut into eight pieces), which are a different pair and are routinely confused with these
• That a product of consecutive integers contains a multiple of 2 and of 3 (What "rational" means, and why the denominator cannot be zero)

What they should be able to do

• Multiply (x − y) by x² + xy + y² and account for every term that cancels
• Explain why the cancellation happens, in terms of how the second bracket is built
• Predict (x + y)(x² − xy + y²) from the first result without multiplying it out
• Show that x⁴ − y⁴ carries x − y among its factors, and argue the same for x⁵ − y⁵
• Multiply (x + y + z) by x² + y² + z² − xy − xz − yz and identify the surviving terms
• Explain why the leftover term is −3xyz rather than −xyz
• Use the three-term identity together with the three-term square to compute a sum of cubes from a sum, a sum of squares and a product
• Factorise a four-term expression of the form a³ + b³ + c³ − 3abc, including cases where a common factor must be removed first
• Prove that n³ − n is divisible by 6 for every natural number n
• Recognise the special case a + b + c = 0 and use it to evaluate an expression in one step

Where it usually goes wrong

• "x³ − y³ = (x − y)³." The two are completely different, and this is the most damaging confusion in the whole chapter because both are in the p. 91 list. Expand both once and leave them side by side.
• "x² + xy + y² is a perfect square, so it must be (x + y)²." (x + y)² has 2xy in the middle. This bracket has xy, and it does not factorise over the rationals. Say so, or students will try to simplify it.
• "The signs in the second bracket are arbitrary." They are forced. The minus in x³ − y³ pairs with the plus in x² + xy + y², and the plus in x³ + y³ pairs with the minus in x² − xy + y². Getting the pairing backwards is the standard examination error.
• "The four terms cancel by luck." They cancel because the second bracket's terms step by one degree at a time, so each product from the x side has a partner from the −y side. Draw the six products and join the pairs.
• "−3xyz must come from one term." It comes from three, one out of each of the three groups of the expansion. That is the entire reason for the 3.
• "You have to find x, y and z before you can find x³ + y³ + z³." Example 15 never does. All the information used is symmetric, and the identity is exactly the tool for converting symmetric information into a sum of cubes.
• "a³ + b³ + c³ = 3abc always." Only when a + b + c = 0. That is what makes Q13 a one-line problem and it is not a general fact.
• "n³ − n needs a divisibility rule." It needs a factorisation. Once it is written as three consecutive integers there is nothing left to prove.

Questions to check understanding

• Factorise a difference or sum of cubes, including cases with coefficients or fractions
• Factorise a four-term expression of the form a³ + b³ + c³ − 3abc, with a common factor removed first if necessary
• Given a sum, a sum of squares and a product of three numbers, find their sum of cubes
• Prove a divisibility statement by factorising into consecutive integers
• Evaluate a cubic expression using the condition that a sum of three quantities is zero
• Show that xⁿ − yⁿ carries x − y among its factors, for a stated small n

Examples worth working on the board

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

• Two products set as an experiment (§4.7, p. 84). The page numbers two products and invites the reader to multiply them: (x − y)(x² + xy + y²) and (x + y)(x² − xy + y²). It then works only the first.
• The first product, worked (§4.7, p. 84). The page prints a six-term expansion — x³, then x²y, then xy², then −x²y, then −xy², and last −y³ — and then reprints the same line with the cancelling terms struck through. The strike-throughs are in the artwork and do not extract, so a text-layer reading sees the line twice with no explanation. Verified: x³ − y³.
• Why the cancellation happens. Not printed; this is added here. Multiplying by x raises every term of the second bracket one degree; multiplying by −y lowers the sign and raises the y-degree. Because the second bracket runs x², xy, y² with a step of one in each direction, the term x times xy lands on the same monomial as −y times x², and likewise for the other pair. Only x times x² and −y times y² have nothing to meet, and they are the two survivors. Drawing the six products in a 2 × 3 grid and joining the pairs is the clearest way to show it.
• The other one, predicted (§4.7, p. 85). The page says the first result is an identity, invites the reader to check it on chosen values, and asks for a prediction of (x + y)(x² − xy + y²). No answer is printed in §4.7. Verified: x³ + y³, and the same identity is listed on p. 91 in the chapter's summary of identities. So the answer is in the volume, five pages later.
• Think and Reflect, powers of a difference (§4.7, p. 85). The box recalls x² − y² = (x − y)(x + y) and the newly found x³ − y³ = (x − y)(x² + xy + y²), observes that x − y is common to both, and asks whether x⁴ − y⁴ carries x − y as a factor too. It supplies the step x⁴ − y⁴ = (x²)² − (y²)² = (x² − y²)(x² + y²), then asks the reader to see how x − y follows, and finally asks about x⁵ − y⁵. No answers are printed. Verified: x⁴ − y⁴ = (x − y)(x + y)(x² + y²); and x⁵ − y⁵ = (x − y)(x⁴ + x³y + x²y² + xy³ + y⁴), where the second bracket is the same construction as before with five terms instead of three. The general reason, which the explanation may state in one line: substituting x = y makes xⁿ − yⁿ zero, and an expression that vanishes when x = y has x − y as a factor.
• The three-letter product (§4.7, p. 85). Multiply (x + y + z) by x² + y² + z² − xy − xz − yz. The page prints the eighteen products in three grouped lines, then reprints them with the cancellations struck through. Verified: x³ + y³ + z³ − 3xyz. The accounting: eighteen products in all; twelve of them cancel in six pairs; three survive as x³, y³ and z³; and each of the three groups contributes exactly one −xyz, which is where the coefficient 3 comes from.
• Example 15 (§4.7, pp. 85–86). Three numbers x, y, z whose sum is 10, whose product is 25, and whose squares total 38. Find x³ + y³ + z³. Inputs: those three values. The page substitutes into the identity to get (10)(38 − (xy + xz + yz)) = x³ + y³ + z³ − 3(25), rearranges to x³ + y³ + z³ = 455 − 10(xy + xz + yz), then turns to the three-term square to get the missing quantity: 100 = 38 + 2(xy + xz + yz), so xy + xz + yz = 31. Verified: 455 − 310 = 145. The point: at no stage is any one of the three numbers found, and none is needed. Every quantity used is symmetric in x, y, z.
• The two identities working together (§4.7–§4.8, p. 86). Section 10 exists because the chapter's own solution needs the earlier identity. The cubic identity supplies the shape of the answer, the square supplies the one unknown quantity in it. That pairing is the section's real lesson.
• Factorising with the three-term identity (End-of-Chapter Q3, p. 89). (vii) p³ + 27q³ + r³ − 9pqr. Verified: with a = p, b = 3q, c = r, the product 3abc is 9pqr, so the factorisation is (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr). (ix) 9x³ − (8/3)y³ + z³/3 + 6xyz. Verified: take 1/3 out first, giving (1/3)(27x³ − 8y³ + z³ + 18xyz), and then with a = 3x, b = −2y, c = z the product 3abc is −18xyz, so the answer is (1/3)(3x − 2y + z)(9x² + 4y² + z² + 6xy + 2yz − 3xz). This is the hardest item in the chapter, and it needs the common-factor move from Exercise Set 4.2 as well as this identity.
• Q11 (End-of-Chapter, p. 90, starred). Given a + b + c = 5 and ab + bc + ca = 10, prove a³ + b³ + c³ − 3abc = −25. Verified: a² + b² + c² = 25 − 20 = 5, so the identity gives 5 × (5 − 10) = −25.
• Q12 (End-of-Chapter, p. 90, starred). Show by factorising that n³ − n is divisible by 6 for every natural number n. Verified: n³ − n = n(n² − 1) = (n − 1)n(n + 1), three consecutive integers; one of any two consecutive integers is even, and one of any three is a multiple of 3, so the product carries both factors and hence 6.
• Q13 (End-of-Chapter, p. 90, starred). (i) Evaluate x³ + y³ − 12xy + 64 when x + y = −4. (ii) Evaluate x³ − 8y³ − 36xy − 216 when x = 2y + 6. Verified: both are 0. For (i), read the expression as x³ + y³ + 4³ − 3(x)(y)(4), so the first factor is x + y + 4, which the condition makes zero. For (ii), read it as x³ + (−2y)³ + (−6)³ − 3(x)(−2y)(−6), so the first factor is x − 2y − 6, which the condition makes zero. The general lesson worth stating: when a + b + c = 0, the identity collapses to a³ + b³ + c³ = 3abc.

Figures to have open

• §4.7's second half prints no figures at all. Everything here is symbolic, which is itself worth noting: after two sections of pictures the chapter finds its most powerful identity with no picture available.
• A 2 × 3 grid of the six products with the cancelling pairs joined. Standard schematic, not in the book, and the single most useful image in the topic.
• A three-column tally for the eighteen-term expansion, showing which column each −xyz comes from. Standard schematic.
• A side-by-side card of x³ − y³ = (x − y)(x² + xy + y²) against (x − y)³ = x³ − 3x²y + 3xy² − y³, to be held in view whenever either is used. Standard schematic.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 4, "Exploring Algebraic Identities": §4.7 "Finding New Identities", pp. 84–86, including Example 15, whose solution finishes on p. 86 at the head of §4.8.
• Think and Reflect, p. 85 — the x⁴ − y⁴ and x⁵ − y⁵ prompts.
• End-of-Chapter Exercises: Q1 (v), p. 88; Q3 (iii), (vi), (vii), (ix), p. 89; Q11, Q12, Q13, p. 90, all three starred.
• Chapter summary, p. 90; the identity list on p. 91 carries x³ − y³, x³ + y³ and the three-term cubic identity.
• Companion topic: Cubes: (a ± b)³ from a cube cut into eight pieces, whose two cube identities must be kept visibly distinct from these.