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Question 12 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)
Last updated at May 18, 2026 by Teachoo
End-of-Chapter Exercises
Class 9
Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)
- Consecutive Square Numbers
- Visualising Algebraic Identities
- Exercise Set 4.1
- Factorisation of Algebraic Expressions Using Identities
- Exercise Set 4.2
- More Identities
- Exercise Set 4.3
- Rewriting a2 - b2 Identity
- Factorisation Using Algebra Tiles
- Factorisation by Splitting the Middle term
- Exercise Set 4.4
- Finding New Algebraic Identities
- Simplifying Rational Expressions
- Exercise Set 4.5
- Word Problems
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End-of-Chapter Exercises
Transcript
Question 12 By factoring the expression, check that 𝑛^3−𝑛 is always divisible by 6 for all natural numbers 𝑛. Give reasons. First, we factorise 𝑛^3−𝑛 Now, 𝒏^𝟑−𝒏=𝑛(𝑛^2−1) =𝑛(𝑛^2−1^2 ) =𝒏(𝒏−𝟏)(𝒏+𝟏) Let's arrange these in order from smallest to largest: (𝒏−𝟏)× 𝒏 × (𝒏+𝟏) Looking at the brackets, they represent three consecutive numbers (like 4, 5, 6 or 11, 12, 13). And, we know that In any three consecutive numbers, at least one of them MUST be an even number (divisible by 2) exactly one of them MUST be a multiple of 3 Now, If a number is multiplied by 2 , and then multiplied by 3 Then, Total product is guaranteed to be a multiple of 𝟐 × 𝟑=𝟔 Thus, our expression 𝑛^3−𝑛 is always divisible by 6 Hence proved
