If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer
[Hint: write aⁿ = (a – b + b)ⁿ and expand]

Solution:

We have to prove that a – b is a factor of aⁿ – bⁿ. 

For this, we have to prove that aⁿ – bⁿ = k (a - b), for some integer k.

an = (a – b + b)n

= [ (a-b) + b ]n

Using the binomial theorem,

an = nC₀ (a-b)n + nC₁ (a-b)n-1 b1+ nC₂ (a-b)n-2 b2 + ... + nCₙ₋₁ (a-b) bn-1 + nCₙ bn

an = (a-b) [nC₀ (a-b)n-1 + nC₁ (a-b)n-2 b1+ nC₂ (a-b)n-2 b2 + ... + nCₙ₋₁ bn-1]+ bn  (because nCₙ = 1)

an - bn = (a-b) [nC₀ (a-b)n-1 + nC₁ (a-b)n-2 b1+ nC₂ (a-b)n-2 b2 + ... + nCₙ₋₁ bn-1] 

an - bn = (a-b) k, where k = nC₀ (a-b)n-1 + nC₁ (a-b)n-2 b1+ nC₂ (a-b)n-2 b2 + ... + nCₙ₋₁ bn-1.

Therefore, a – b is a factor of aⁿ – bⁿ

NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 4

If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer. [Hint: write aⁿ = (a – b + b)ⁿ and expand]

Summary:

If a and b are distinct integers, we proved that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer

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