Mathematics Class 9 — Ganita…
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Trinomial and Cube Identities

🎓 Class 9 Mathematics CBSE Theory Ch 4 — Exploring Algebraic Identities ⏱ ~45 min
🌐 Language: [gtranslate]

This MCQ module is based on: Trinomial and Cube Identities

This mathematics assessment will be based on: Trinomial and Cube Identities
Targeting Class 9 level in Algebra, with Intermediate difficulty.

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4.5 The Identity \((x+a)(x+b) = x^2 + (a+b)x + ab\)

What if we want to multiply two binomials whose first terms match but whose second terms differ? Multiply directly: \[ (x+a)(x+b) = x \cdot x + x \cdot b + a \cdot x + a \cdot b = x^2 + (a+b)x + ab. \]

Identity IV: \((x+a)(x+b) = x^2 + (a+b)\,x + ab\)

This identity is extremely useful both for multiplying two-digit numbers near a round number and for factorising quadratics.

Worked Examples

Example 11. Compute \(103 \times 107\).
Take \(x = 100, a = 3, b = 7\). Then \(103 \times 107 = 100^2 + (3+7)(100) + (3 \cdot 7) = 10000 + 1000 + 21 = 11021\).
Example 12. Expand \((y + 5)(y + 12)\).
\(a + b = 17,\ ab = 60\). \((y+5)(y+12) = y^2 + 17y + 60\).
Example 13. Factorise \(x^2 + 10x + 21\).
Find two numbers \(a, b\) with \(a + b = 10\) and \(ab = 21\). Try \(a=3, b=7\): sum 10 ✓, product 21 ✓. Hence \(x^2+10x+21 = (x+3)(x+7)\).
Example 14. Factorise \(y^2 - 5y - 24\).
Need \(a + b = -5,\ ab = -24\). Try \(a = -8, b = 3\): \(-8 + 3 = -5\) ✓, \(-8 \cdot 3 = -24\) ✓. So \(y^2 - 5y - 24 = (y - 8)(y + 3)\).

4.6 Squaring a Trinomial: \((a+b+c)^2\)

What if three quantities are added before squaring? We may discover the formula by treating \(a + b\) as a single bundle: \[ (a+b+c)^2 = ((a+b) + c)^2 = (a+b)^2 + 2(a+b)c + c^2. \] Expanding using Identity I: \( = a^2 + 2ab + b^2 + 2ac + 2bc + c^2.\)

Identity V: \((a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)

Geometric Picture — A 3 × 3 Grid

A square of side \((a+b+c)\) splits into a 3 × 3 grid of nine pieces: three squares (\(a^2, b^2, c^2\)) on the diagonal and six off-diagonal rectangles. Pairing equal off-diagonal rectangles gives \(2ab, 2bc, 2ca\). Total: \(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\). ✓

a² b² c² ab ab ac ac bc bc
Fig 4.4 — \((a+b+c)^2\) splits into 3 squares + 6 rectangles = \(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\).

Worked Examples

Example 15. Expand \((2x + 3y + 4z)^2\).
Apply Identity V with \(a=2x, b=3y, c=4z\):
\((2x)^2 + (3y)^2 + (4z)^2 + 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x)\)
= \(4x^2 + 9y^2 + 16z^2 + 12xy + 24yz + 16zx\).
Example 16. If \(a + b + c = 9\) and \(ab + bc + ca = 23\), find \(a^2 + b^2 + c^2\).
By Identity V: \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\). So \(81 = a^2 + b^2 + c^2 + 2(23) = a^2 + b^2 + c^2 + 46\). Hence \(a^2 + b^2 + c^2 = 35\).

4.7 Cube Identities — \((a \pm b)^3\)

To cube a sum, multiply \((a+b)\) by \((a+b)^2\): \[(a+b)^3 = (a+b)(a+b)^2 = (a+b)(a^2 + 2ab + b^2)\] \[= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 = a^3 + 3a^2b + 3ab^2 + b^3.\]

Identity VI: \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a+b)\)

Similarly, replacing \(b\) with \(-b\) gives:

Identity VII: \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a-b)\)

Geometric View — A Cube of Side (a+b)

A cube of side \((a+b)\) has volume \((a+b)^3\). Slicing parallel to its faces produces 8 sub-pieces:
• 1 cube of side \(a\), volume \(a^3\)
• 1 cube of side \(b\), volume \(b^3\)
• 3 rectangular slabs of dimensions \(a \times a \times b\), each volume \(a^2 b\) → total \(3a^2 b\)
• 3 rectangular slabs of dimensions \(a \times b \times b\), each volume \(ab^2\) → total \(3ab^2\)
Sum: \(a^3 + 3a^2b + 3ab^2 + b^3\). ✓

a³ a²b a²b ab² cube of side (a+b) a b a b 8 pieces: 1·a³ + 1·b³ + 3·a²b + 3·ab² (b³ hidden behind)
Fig 4.5 — Volume decomposition of \((a+b)^3\) — front face shows visible pieces; 8 pieces in total.

Worked Examples — Cubes

Example 17. Compute \((102)^3\).
Take \(a = 100, b = 2\). \((100+2)^3 = 100^3 + 3(100^2)(2) + 3(100)(4) + 8 = 1000000 + 60000 + 1200 + 8 = 1061208\).
Example 18. Expand \((3x - 2y)^3\).
\(a = 3x, b = 2y\): \((3x)^3 - 3(3x)^2(2y) + 3(3x)(2y)^2 - (2y)^3 = 27x^3 - 54x^2y + 36xy^2 - 8y^3\).
Example 19. If \(x + \frac{1}{x} = 4\), find \(x^3 + \frac{1}{x^3}\).
Cube both sides of \(x + \frac{1}{x} = 4\) and use \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\):
\(\left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x \cdot \frac{1}{x}\right)\left(x + \frac{1}{x}\right) = x^3 + \frac{1}{x^3} + 3(1)(4) = x^3 + \frac{1}{x^3} + 12\).
LHS = \(64\). Hence \(x^3 + \frac{1}{x^3} = 64 - 12 = 52\).

4.8 Sum and Difference of Cubes

Two more powerful factorisation identities:

Identity VIII: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
Identity IX: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

Verification of VIII: expand RHS: \[ (a+b)(a^2 - ab + b^2) = a^3 - a^2b + ab^2 + a^2 b - ab^2 + b^3 = a^3 + b^3.\ \checkmark \]

A Beautiful Three-Variable Identity

Identity X
\[a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\] A lovely consequence: if \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).

Worked Examples — Sum/Difference of Cubes

Example 20. Factorise \(8x^3 + 27y^3\).
\(8x^3 = (2x)^3,\ 27y^3 = (3y)^3\). By Identity VIII: \((2x + 3y)((2x)^2 - (2x)(3y) + (3y)^2) = (2x+3y)(4x^2 - 6xy + 9y^2)\).
Example 21. Factorise \(125 - 8a^3\).
\(125 = 5^3,\ 8a^3 = (2a)^3\). By Identity IX: \(5^3 - (2a)^3 = (5 - 2a)(25 + 10a + 4a^2)\).
Example 22. If \(a + b + c = 0\), evaluate \(\frac{a^3 + b^3 + c^3}{abc}\).
From Identity X corollary: \(a^3 + b^3 + c^3 = 3abc\). Therefore \(\frac{a^3 + b^3 + c^3}{abc} = \frac{3abc}{abc} = 3\).
Activity: The "Magic" Number Trick
L4 Analyse
Materials: Pen, paper, friend.
Predict: Multiplying any two numbers \(a\) and \(b\) where \(a+b = 10\) gives a number ending in the digits of \(a \cdot b\) padded into "tens × tens × 100 + product". Why does this trick work?
  1. Pick numbers like \(67 \times 63\). Note that the tens digit is the same (6) and the units digits sum to 10 (7 + 3 = 10).
  2. Method: tens digit \(\times\) (tens digit + 1) gives the leading digits. Here: \(6 \times 7 = 42\). The trailing digits = product of units = \(7 \times 3 = 21\). Answer: \(\mathbf{4221}\).
  3. Verify by long multiplication: \(67 \times 63 = 4221\). ✓
  4. Try \(82 \times 88\): leading = \(8 \times 9 = 72\); trailing = \(2 \times 8 = 16\). Answer: \(\mathbf{7216}\). ✓

The proof uses Identity IV. Let the two numbers be \((10n + a)\) and \((10n + b)\) with \(a + b = 10\). Multiply:

\((10n+a)(10n+b) = 100n^2 + 10n(a+b) + ab = 100n^2 + 100n + ab = 100 \cdot n(n+1) + ab.\)

This is exactly: \(n(n+1)\) followed by the two-digit \(ab\) — which is the trick!

Competency-Based Questions

Scenario: A wooden cube of side \((x + 3)\) cm is being cut by a carpenter. From it she removes a smaller cube of side \(x\) cm at one corner. The leftover wood will be used as filler for crafts.
Q1. Compute the volume of leftover wood, expressed in expanded polynomial form.
L3 Apply
  • (a) \(9x^2 + 27x + 27\)
  • (b) \(9x^2 + 27\)
  • (c) \(3x^3 + 27\)
  • (d) \(27\)
Answer (a). Leftover = \((x+3)^3 - x^3 = (x^3 + 9x^2 + 27x + 27) - x^3 = 9x^2 + 27x + 27\) cm³.
Q2. Verify your answer using the identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) with \(a = x+3, b = x\). Show all steps.
L4 Analyse
\(a - b = 3\); \(a^2 = x^2 + 6x + 9\); \(ab = x(x+3) = x^2 + 3x\); \(b^2 = x^2\). Sum: \(a^2 + ab + b^2 = 3x^2 + 9x + 9\). Multiply by \((a-b) = 3\): \(9x^2 + 27x + 27\). ✓ Matches Q1.
Q3. The carpenter says, "If x doubles, the leftover volume more than doubles." Evaluate this claim mathematically using \(x = 5\) vs \(x = 10\).
L5 Evaluate
At \(x = 5\): \(9(25) + 27(5) + 27 = 225 + 135 + 27 = 387\) cm³. At \(x = 10\): \(9(100) + 27(10) + 27 = 900 + 270 + 27 = 1197\) cm³. Ratio = \(1197 / 387 \approx 3.09\) — yes, more than doubled, in fact tripled. The carpenter's claim is true; the leading \(9x^2\) term dominates and grows quadratically.
Q4. Design a wooden block (any shape) whose volume equals the leftover volume \(9x^2 + 27x + 27\) and that the carpenter can describe with simple integer dimensions when \(x = 6\). Specify dimensions and verify.
L6 Create
At \(x=6\): volume = \(9(36) + 27(6) + 27 = 324 + 162 + 27 = 513\) cm³. Notice \(513 = 3 \times 171 = 3 \times 3 \times 57 = 27 \times 19\). So one design: a cuboid of \(3 \times 9 \times 19\) cm. Verify: \(3 \cdot 9 \cdot 19 = 513\) ✓. Alternative: \(9 \times 9 \times \frac{513}{81}\) — non-integer. The \(3 \times 9 \times 19\) cuboid works. Also: factoring \(9x^2 + 27x + 27 = 9(x^2 + 3x + 3)\), so for any \(x\), one dimension can be 9 cm and the cross-section area \((x^2+3x+3)\) cm² — useful general design.

Assertion–Reason Questions

Assertion (A): \(x^2 - 7x + 12 = (x - 3)(x - 4)\).
Reason (R): The numbers \(-3\) and \(-4\) satisfy \(a + b = -7\) and \(ab = 12\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) — \(-3 + (-4) = -7\) ✓ and \((-3)(-4) = 12\) ✓. R correctly justifies the factorisation via Identity IV.
Assertion (A): If \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).
Reason (R): \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) — Setting \((a+b+c) = 0\) in R kills the RHS, giving exactly \(a^3+b^3+c^3 = 3abc\). R explains A directly.
Assertion (A): \((a+b+c)^2 = a^2 + b^2 + c^2\).
Reason (R): Squaring distributes over addition.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(d) — A is FALSE (missing the cross-terms \(2ab+2bc+2ca\)). R is also FALSE: squaring does NOT distribute over addition; that's the very mistake the assertion makes. Wait — strictly R says squaring distributes (false). So technically (d) requires R true. Best answer: both A and R are false, but in the standard four-option ARQ scheme, mark the option where A is false. Many boards accept this as (d) treating R as the "rule" being incorrectly stated. Read carefully — primary point: A is wrong because of the cross-terms.

Frequently Asked Questions

What is the expansion of (a + b + c)^2 in Class 9 Maths?

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca. Square each variable and add twice the product of every pair. For instance, (x + 2y + 3z)^2 = x^2 + 4y^2 + 9z^2 + 4xy + 12yz + 6xz.

What are the cube identities (a + b)^3 and (a - b)^3?

(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 and (a - b)^3 = a^3 - 3a^2 b + 3ab^2 - b^3. They can also be written as (a + b)^3 = a^3 + b^3 + 3ab(a + b) and (a - b)^3 = a^3 - b^3 - 3ab(a - b).

How do you factorise a^3 + b^3 and a^3 - b^3?

Use the identities a^3 + b^3 = (a + b)(a^2 - ab + b^2) and a^3 - b^3 = (a - b)(a^2 + ab + b^2). For example, x^3 + 27 = x^3 + 3^3 = (x + 3)(x^2 - 3x + 9), and 8 - y^3 = 2^3 - y^3 = (2 - y)(4 + 2y + y^2).

If a + b + c = 0, what is the value of a^3 + b^3 + c^3?

There is a beautiful identity: when a + b + c = 0, then a^3 + b^3 + c^3 = 3abc. This follows from the general identity a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).

How do you compute 103^3 using the cube identity?

Write 103 = 100 + 3. Then 103^3 = (100 + 3)^3 = 100^3 + 3(100)^2(3) + 3(100)(3)^2 + 3^3 = 1,000,000 + 90,000 + 2700 + 27 = 1,092,727.

How is the identity a^3 + b^3 + c^3 - 3abc factored?

It factorises as (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca). This is widely used in olympiad-style and HOTS questions and immediately implies that a^3 + b^3 + c^3 = 3abc whenever a + b + c = 0.

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