NCERT Solutions for Class 8 Maths Chapter 6: We Distribute, Yet Things Multiply – Free PDF Download

Chapter 6 — We Distribute, Yet Things Multiply — extends the distributive property of multiplication over addition to algebra: expanding products of binomials and multi-term expressions term by term, and building three key identities — (a+b)² = a²+2ab+b², (a−b)² = a²−2ab+b², and (a+b)(a−b) = a²−b² — for fast mental arithmetic, number puzzles, and geometric area problems. The chapter has three Figure It Out blocks totalling 20 questions. Below are complete, verified answers, sourced directly from the official NCERT PDF and independently recalculated. These Class 8 Mathematics Chapter 6 solutions are also useful as quick revision notes before exams.

NCERT Solutions for Class 8 Maths Chapter 6: We Distribute, Yet Things Multiply

Figure It Out — Section 6.1: Some Properties of Multiplication (5 Questions)

1. In a multiplication grid, the middle number of a 3×3 frame is pq (columns headed p−1, p, p+1; rows headed q−1, q, q+1). Write expressions for the other 8 numbers in the grid.
Top row: (p−1)(q−1), p(q−1), (p+1)(q−1). Middle row: (p−1)q, pq, (p+1)q. Bottom row: (p−1)(q+1), p(q+1), (p+1)(q+1).

2. Expand the following products: (i) (3+u)(v−3) (ii) ⅔(15+6a) (iii) (10a+b)(10c+d) (iv) (3−x)(x−6) (v) (−5a+b)(c+d) (vi) (5+z)(y+9).
(i) 3v−9+uv−3u. (ii) 10+4a. (iii) 100ac+10ad+10bc+bd. (iv) −x²+9x−18. (v) −5ac−5ad+bc+bd. (vi) 5y+45+yz+9z.

3. Find 3 examples where the product of two numbers stays the same when one is increased by 2 and the other decreased by 4.
Setting ab=(a+2)(b−4) gives b=2a+4. So (1,6), (2,8), (3,10) all work — e.g. 1×6=6 and 3×2=6.

4. Expand (i) (a+ab−3b²)(4+b) (ii) (4y+7)(y+11z−3).
(i) 4a+5ab−12b²+ab²−3b³. (ii) 4y²+44yz−5y+77z−21.

5. Expand (a−b)(a+b), (a−b)(a²+ab+b²), and (a−b)(a³+a²b+ab²+b³). What pattern emerges, and what’s the next identity?
They simplify to a²−b², a³−b³, and a⁴−b⁴ respectively. Pattern: (a−b)(aⁿ+aⁿ⁻¹b+…+bⁿ) = aⁿ⁺¹−bⁿ⁺¹. Next: (a−b)(a⁴+a³b+a²b²+ab³+b⁴) = a⁵−b⁵, verified by expansion.

Figure It Out — Section 6.2: Special Cases of the Distributive Property (4 Questions)

1. Which is greater: (a−b)² or (b−a)²? Justify.
They are equal — both expand to a²+b²−2ab, since squaring removes the sign of the difference.

2. Express 100 as the difference of two squares.
Using a²−b²=(a+b)(a−b)=100 with a+b=50, a−b=2 gives a=26, b=24. 26²−24² = 676−576 = 100.

3. Find 406², 72², 145², 1097², and 124² using the identities learned.
406²=(400+6)²=164836. 72²=(50+22)²=5184. 145²=(150−5)²=21025. 1097²=(1100−3)²=1203409. 124²=(100+24)²=15376.

4. Do 2(a²+b²)=(a+b)²+(a−b)² and a²−b²=(a+b)(a−b) hold only for counting numbers, or also for negative integers and fractions?
Both identities hold for all three cases — counting numbers, negative integers, and fractions — because they are proved by pure algebraic expansion, which never assumes a sign or type of number. E.g. with a=−4, b=−2: LHS=2(16+4)=40, RHS=(−6)²+(−2)²=40.

Figure It Out — Sections 6.3 & 6.4: Mind the Mistake / All Ways Lead to the Bay (11 Questions)

1. Compute using the suggested identity: (i) 46² (ii) 397×403 (iii) 91² (iv) 43×45.
(i) 46²=(40+6)²=2116. (ii) 397×403=(400−3)(400+3)=400²−3²=159991. (iii) 91²=(100−9)²=8281. (iv) 43×45=(44−1)(44+1)=44²−1=1935.

2. Use an identity or the distributive property to find: (i) (p−1)(p+11) (ii) (3a−9b)(3a+9b) (iii) −(2y+5)(3y+4) (iv) (6x+5y)² (v) (2x−½)² (vi) 7p×3r×(p+2).
(i) p²+10p−11. (ii) 9a²−81b². (iii) −6y²−23y−20. (iv) 36x²+60xy+25y². (v) 4x²−2x+¼. (vi) 21p²r+42pr.

3. Identify the correct algebraic expression: (i) two more than a square number (ii) sum of squares of two consecutive numbers.
(i) s²+2. (ii) m²+(m+1)².

4. In a 2×2 square of calendar dates, find the products along each diagonal. What do you notice, and why?
The two diagonal products always differ by 7. Labelling the square n, n+1 (top) and n+7, n+8 (bottom), the diagonals give n(n+8)=n²+8n and (n+1)(n+7)=n²+8n+7 — a difference of exactly 7, since calendar rows are 7 apart.

5. Verify which statements are true: (i) (k+1)(k+2)−(k+3) is always a multiple of 2 (ii) (2q+1)(2q−3) is a multiple of 4 (iii) squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than a multiple of 8 (iv) (6n+2)²−(4n+3)² is 5 less than a square number.
(i) Expands to k²+2k−1, which is not always even (k=1 gives 2, even; k=2 gives 7, odd) — false. (ii) Expands to 4q²−4q−3, which is always odd — false. (iii) True for both parts (e.g. 6²=36=4×9; 7²=49=8×6+1). (iv) Simplifies to 20n²−16n−5, which is not generally 5 less than a perfect square — false in general.

6. A number leaves remainder 3 when divided by 7, another leaves remainder 5. Find the remainder when their sum, difference, and product are divided by 7.
With x=7a+3, y=7b+5: sum’s remainder is 1 (3+5=8=7+1), difference’s remainder is 5 (since −2≡5 mod 7), and product’s remainder is 1 (3×5=15=14+1).

7. Square the middle of three consecutive numbers and subtract the product of the outer two. What pattern do you get, and how do you prove it algebraically?
The result is always 1 (e.g. 8²−7×9=64−63=1). Algebraically, with numbers a−1, a, a+1: a²−(a+1)(a−1) = a²−(a²−1) = 1.

8. Add two numbers, then multiply by half their sum. Show this equals half the square of their sum.
(a+b)×½(a+b) = ½(a+b)² — true by direct simplification.

9. Which is larger, without fully computing: (i) 14×26 or 16×24 (ii) 25×75 or 26×74?
(i) Writing 16×24 = (14+2)(26−2) = 14×26 + 2×10, so 16×24 is larger by 20. (ii) Writing 26×74 = (25+1)(75−1) = 25×75 + 49, so 26×74 is larger by 49.

10. A park has two square green plots of area g² sq ft each, surrounded by a walking path w ft wide. Find the area to be tiled.
Overall park area = (4w+2g)(2w+g) = 8w²+8wg+2g². Subtracting the 2g² green area leaves (8w²+8wg) sq ft to be tiled.

11. For each growing pattern of unit squares: how many units are in Step 10, and what’s the expression for Step y?
Pattern (a): Step y has (y+2)² unit squares, so Step 10 has 12² = 144. Pattern (b): Step y has (y+1)²+y unit squares, so Step 10 has 11²+10 = 131.

Why This Chapter Matters

This chapter turns the distributive property into a toolkit — three reusable identities that make mental arithmetic, algebraic simplification, and later factorisation dramatically faster. Recognising when a product fits (a+b)², (a−b)², or (a+b)(a−b) is a skill used throughout higher algebra, from Class 9 polynomial factorisation to competitive-exam shortcuts.

More on This Chapter

Extra Questions (HOTS) | Revision Notes | Formulas Handbook | Class 8 Maths Book

Frequently Asked Questions

Do I need to memorise all three identities separately, or can I derive them?
Both are useful: memorise them for speed, but you should also be able to derive each one from the distributive property a(b+c)=ab+ac in seconds — that’s what makes them reliable under exam pressure instead of easy to misremember.

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.