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MathematicsClass 9Exploring Algebraic

Exploring Algebraic | Class 9 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 5 min read

Exploring Algebraic | Class 9 Mathematics Notes

Exploring Algebraic – this guide gives you a concise, exam-ready overview of Exploring Algebraic from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

4.7 FINDING NEW IDENTITIES

This section explores the derivation and visualization of new algebraic identities, especially involving cubes. Starting with the cube of a binomial (a + b)³, the identity is derived using distributive property as (a + b)³ = a³ + 3a²b + 3ab² + b³. This is visualised geometrically by dividing a cube of edge (a + b) into smaller cubes and cuboids, corresponding to the terms in the expansion. The section also derives the identity for (a - b)³ by substituting b with -b, resulting in (a - b)³ = a³ - 3a²b + 3ab² - b³, noting the alternating signs. Examples demonstrate how to identify the side of a cube given its volume expressed as a sum of terms matching the identity. Further, the section explores the factorisation of differences and sums of cubes, deriving identities such as (x - y)(x² + xy + y²) = x³ - y³ and (x + y)(x² - xy + y²) = x³ + y³. The section culminates with the identity involving three variables: (x + y + z)(x² + y² + z² - xy - yz - zx) = x³ + y³ + z³ - 3xyz, and applies it to find the sum of cubes given sums and products of variables. This exploration deepens understanding of algebraic structures and their applications.

📊 Diagram: Fig. 4.9: A cube of edge $a + b$; Fig. 4.10: Representation of the identity $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

🧪 Activity: Students are encouraged to verify identities by substituting values and to explore patterns in powers and factorisation.

🔗 Connection: Leads to simplifying rational expressions using factorisation.

Frequently asked questions

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i) 117^2 (ii) 78^2 (iii) 198^2 (iv) 214^2 (v) 1104^2 (vi) 1120^2

Solutions:

(i) 117^2 Use (a + b)^2 = a^2 + 2ab + b^2 117 = 100 + 17 So, 117^2 = 100^2 + 210017 + 17^2 = 10000 + 3400 + 289 = 13689

(ii) 78^2 Use (a - b)^2 = a^2 - 2ab + b^2 78 = 80 - 2 So, 78^2 = 80^2 - 2802 + 2^2 = 6400 - 320 + 4 = 6084

(iii) 198^2 Use (a - b)^2 198 = 200 - 2 198^2 = 200^2 - 22002 + 2^2 = 40000 - 800 + 4 = 39204

(iv) 214^2 Use (a + b)^2 214 = 200 + 14 214^2 = 200^2 + 220014 + 14^2 = 40000 + 5600 + 196 = 45796

(v) 1104^2 Use (a + b)^2 1104 = 1100 + 4 1104^2 = 1100^2

2. Factor using suitable identities: (i) 16y^{2} - 24y + 9 (ii) (9/4) s^2 + 6st + 4t^2 (iii) (m^2)/9 + (mk)/3 + (k^2)/4 + 3nk + 2mn + 9n^2 (iv) (p^2)/16 - 2 + 16/(p^2) (v) 9a^{2} + 4b^{2} + c^{2} - 12ab + 6ac - 4bc

Solutions:

(i) 16y^2 - 24y + 9 = (4y)^2 - 24y3 + 3^2 = (4y - 3)^2

(ii) (9/4)s^2 + 6st + 4t^2 = (3s/2)^2 + 2(3s/2)2t + (2t)^2 = (3s/2 + 2t)^2

(iii) (m^2)/9 + (mk)/3 + (k^2)/4 + 3nk + 2mn + 9n^2 Group terms: = (m^2)/9 + (mk)/3 + (k^2)/4 + 2mn + 3nk + 9n^2 Try to write as square of (m/3 + k/2 + 3n)^2 Calculate: (m/3)^2 + (k/2)^2 + (3n)^2 + 2(m/3)(k/2) + 2(k/2)(3n) + 2(m/3)(3n) = (m^2)/9 + (k^2)/4 + 9n^2 + (mk)/3 + 3nk + 2mn Matches given expression exactly. So factorization is (m/3 + k

3. Expand the following using the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca: (i) (p + 3q + 7r)^2 (ii) (3x - 2y + 4z)^2

Solutions:

(i) (p + 3q + 7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2p3q + 23q7r + 27rp = p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp

(ii) (3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 23x(-2y) + 2(-2y)4z + 24z3x = 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx

4. Is this an identity? (a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2.

Yes, this is an identity.

Proof: Expand each square:

(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2ac - 2bc (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc (a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc

Add all three: = (a^2 + b^2 + c^2 + 2ab - 2ac - 2bc) + (a^2 + b^2 + c^2 - 2ab + 2ac - 2bc) + (a^2 + b^2 + c^2 - 2ab - 2ac + 2bc) = 3a^2 + 3b^2 + 3c^2 + (2ab - 2ab - 2ab) + (-2ac + 2ac - 2ac) + (-2bc - 2bc + 2bc) = 3a^2 + 3b^2 + 3c^2 - 2ab - 2ac - 2bc But carefully summing the cross terms: 2ab

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