Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 4 · Chapter companion part
Expand and recognise cube structures, derive sum/difference factors and evaluate reciprocal powers with explicit conditions.
Class 8 English Version: part B of the chapter 4 companion. Eight sections and 30 written tasks. Selected companion coverage, not complete textbook coverage; use all four parts and compare the mapped source methods. The software checks stated numerical values or coefficients; symbolic work, equivalent forms and proofs need guardian or teacher review.
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(a + b)³ = a³ + 3a²b + 3ab² + b³. Each mixed type occurs three times when choosing terms from three factors. For (2x + 3y)³ the coefficients are 8, 36, 54 and 27. The exploded positive-length cube with a = 3 and b = 1 has volumes 27, three 9s, three 3s and 1, totalling 64.
(a − b)³ = a³ − 3a²b + 3ab² − b³. Substitute the full second term: (3x − 2y)³ = 27x³ − 54x²y + 36xy² − 8y³. Cubing a negative entire expression changes its sign, unlike squaring. A physical block model needs positive lengths, while the distributed identity holds for signed real values.
For any expressions U and V, U³ + 3U²V + 3UV² + V³ = (U + V)³. If U = 2m + n and V = m − n, their sum is 3m and the result is 27m³. Similarly U³ − V³ − 3UV(U − V) = (U − V)³. Define and recover the whole substituted expressions, including signs.
For (a + b + c)³, first set U = a + b and cube U + c. Then expand U² and U³. There are three pure cubes, six mixed terms of type a²b with coefficient 3, and the term 6abc. The compact equivalent is a³ + b³ + c³ + 3(a + b)(b + c)(c + a). A negative term is substituted as a signed value throughout; no pair term may be dropped.
a³ + b³ = (a + b)(a² − ab + b²), while a³ − b³ = (a − b)(a² + ab + b²). Verify by multiplying: the unwanted mixed terms cancel. Thus (3x + 2)(9x² − 6x + 4) = 27x³ + 8. A sum of cubes is not generally just the cube of a sum.
If s = a + b and p = ab, then a³ + b³ = s³ − 3ps. If d = a − b, then a³ − b³ = d³ + 3pd. For 8t³ + 36t² + 54t + 25, recognise (2t + 3)³ − 2, because that cube has constant 27. At t = 2, the value is 341. Keep the correction constant rather than declaring a near-cube to be an exact cube.
For x ≠ 0, let s = x + 1/x and t = x − 1/x. Then x³ + 1/x³ = s³ − 3s and x³ − 1/x³ = t³ + 3t. Also x⁴ + 1/x⁴ = t⁴ + 4t² + 2, giving x⁵ − 1/x⁵ = t(x⁴ + x² + 1 + x⁻² + x⁻⁴) = t⁵ + 5t³ + 5t. These follow by multiplying and cancelling; keep denominators nonzero.
Source exercise 4.4 item 6(iii) is false as written: a³ + b³ equals (a + b)³ minus 3ab(a + b), not plus. Exercise 4.4 item 20C also omits z in the cross term. If z = 3x + 5y, then z³ = 27x³ + 125y³ + 45xyz. At x = y = 1, z = 8, the printed left side without z gives 197, but z³ is 512. The corrected cross term gives 360 and restores 512.
For a + b = 9 and ab = 20, the cube sum is 9³ − 3 × 20 × 9 = 189. For example a = 4 and b = 5 gives 64 + 125 = 189. The substituted values check this case; the derived identity supplies the general reason.
Eight original blocks have volumes 27,9,9,9,3,3,3,1 for a = 3, b = 1. Exploded-view gaps are display spacing, not part of the volume.
The false plus sign in exercise 4.4 item 6(iii) belongs to an assertion-selection question. The missing z in item 20C instead makes the requested proof false as written; the lesson distinguishes these cases.
This inventory comes from distributing three binomial factors.
| Chosen type | How many choices | Term |
|---|---|---|
| Three a terms | 1 | a³ |
| Two a terms and one b | 3 | 3a²b |
| One a term and two b | 3 | 3ab² |
| Three b terms | 1 | b³ |
With a = 3, b = 1, the eight block volumes are 27,9,9,9,3,3,3,1. Their sum is 64, agreeing with 4³. For a signed b, algebraic signs change even though physical blocks cannot have negative volume.
Distinguish the cube of a sum from the sum of cubes.
(a + b)³ = a³ + 3a²b + 3ab² + b³ (a − b)³ = a³ − 3a²b + 3ab² − b³ a³ + b³ = (a + b)³ − 3ab(a + b) a³ − b³ = (a − b)³ + 3ab(a − b) a³ + b³ = (a + b)(a² − ab + b²) a³ − b³ = (a − b)(a² + ab + b²)
The minus sign in the sum-of-cubes corollary is essential. The source’s plus-sign assertion is an intentional false statement to diagnose.
Define U = 2p + q, then expand (U − r)³.
(U − r)³ = U³ − 3U²r + 3Ur² − r³ U = 2p + q Result = 8p³ + q³ − r³ + 12p²q − 12p²r + 6pq² + 6pr² − 3q²r + 3qr² − 12pqr
The coefficient of pqr is −12 here because the first term is 2p and the third is −r. A different signed trinomial needs its own substitution.
Every reciprocal requires a nonzero denominator.
s = x + 1/x; t = x − 1/x x³ + 1/x³ = s³ − 3s x³ − 1/x³ = t³ + 3t x⁴ + 1/x⁴ = t⁴ + 4t² + 2 x⁵ − 1/x⁵ = t⁵ + 5t³ + 5t
Retain the central 1 in the factored fifth-power sum. The source creative fifth-power question uses this same derivation type.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original projected blocks with dimensions labelled and a common coordinate scale. Blocks are separated to reveal their faces; display gaps do not add volume. Positive lengths are assumed.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original symbolic relationship map, not a physical area or volume scale. Algebraic identities apply to all permitted real values.
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1728
(10 + 2)³ = 1000 + 600 + 120 + 8 = 1728.
6859
(20 − 1)³ = 8000 − 1200 + 60 − 1 = 6859.
36
3 × 2² × 3 = 36.
54
3 × 2 × 3² = 54.
-54
3 × 3² × (−2) = −54.
-8
(−2y)³ gives −8y³.
91
7³ − 3 × 12 × 7 = 91.
117
3³ + 3 × 10 × 3 = 117.
52
4³ − 3 × 4 = 52.
14
2³ + 3 × 2 = 14.
82
2⁵ + 5 × 2³ + 5 × 2 = 82.
194
The square sum is 14; its square minus 2 is 194.
35
The product is x³ + 8 = 27 + 8 = 35.
37
The product is 8x³ − 27 = 64 − 27 = 37.
341
(2t + 3)³ − 2 = 7³ − 2 = 341.
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8x³ + 36x²y + 54xy² + 27y³. There are three choices for the position of the single different term in each mixed type.
Show either distribution or the cube identity with complete signed terms.
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27x³ − 54x²y + 36xy² − 8y³.
The middle signs alternate; the squared second term is positive while its cube is negative.
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8p³ + q³ − r³ + 12p²q − 12p²r + 6pq² + 6pr² − 3q²r + 3qr² − 12pqr.
Require all ten term types and check the three-variable term coefficient.
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The expression is (U + V)³ = (3m)³ = 27m³.
Recover both substituted expressions before cubing their sum.
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It is (U − V)³ = (a + b)³ = a³ + 3a²b + 3ab² + b³.
Confirm that U − V = a + b, including the double negative.
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(3x + 2)(9x² − 6x + 4). The mixed terms cancel to leave 27x³ + 8.
The quadratic has the opposite middle sign from the linear sum.
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(2p − 3q)(4p² + 6pq + 9q²).
Keep the positive middle term in the quadratic and cube both coefficients.
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The first two factors give x³ + 8, so the product is (x³ + 8)(x³ − 8) = x⁶ − 64.
Apply the cube identity and then the difference-of-squares identity.
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Cube the sum: 64 = x³ + 1/x³ + 3(x + 1/x). Subtract 12 to get 52.
Use x × 1/x = 1 and show why the mixed terms combine.
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Factor the fifth-power difference as t times (x⁴ + x² + 1 + x⁻² + x⁻⁴). The square sum is t² + 2 and fourth sum is t⁴ + 4t² + 2. Therefore the result is t⁵ + 5t³ + 5t, giving 82 at t = 2.
Retain the central 1 and explain both reciprocal sums.
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z³ = 8x³ + 27y³ + 18xy(2x + 3y) = 8x³ + 27y³ + 18xyz.
The third variable remains in the mixed term after substitution.
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Then z = 8. The printed left side is 197, while the right side is 512. The missing factor is z: 27 + 125 + 45 × 8 = 512.
A counterexample disproves the original; show the corrected general identity too.
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The correct sign is minus. At a = b = 1 the asserted right side is 14, while the left side is 2; the corrected right side is 8 − 6 = 2.
Identify this as an assertion to reject, not a formula to memorise.
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Cube a² + b²: c⁶ = a⁶ + b⁶ + 3a²b²(a² + b²). Replace the parenthesis by c².
Do not replace c² by c or cube each term without the mixed contribution.
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The cube constant is 27, so the expression is (2t + 3)³ − 2. At t = 2 it is 341; direct evaluation gives 64 + 144 + 108 + 25 = 341.
Check the constant separately and show both routes.
Make a cube-term inventory and write one composite, one factor and one reciprocal derivation.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 4, printed pages 48–76. It does not replace the complete chapter or its practical teaching.
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