Find (x + 1)⁶ + (x - 1)⁶. Hence or otherwise evaluate (√2 + 1)⁶ + (√2 - 1)⁶

Solution:

Using binomial theorem, the expressions (x + 1)⁶ and (x - 1)⁶ can be expanded as

• (x + 1)⁶ = ⁶C₀ x⁶ + ⁶C₁ x⁵ + ⁶C₂ x⁴ + ⁶C₃ x³ + ⁶C₄ x² + ⁶C₅ x + ⁶C₆
   
• (x - 1)⁶ = ⁶C₀ x⁶ - ⁶C₁ x⁵ + ⁶C₂ x⁴ - ⁶C₃ x³ + ⁶C₄ x² - ⁶C₅ x + ⁶C₆

Therefore,

(x + 1)⁶ + (x - 1)⁶ = [(⁶C₀ x⁶ + ⁶C₁ x⁵ + ⁶C₂ x⁴ + ⁶C₃ x³ + ⁶C₄ x² + ⁶C₅ x + ⁶C₆) + (⁶C₀ x⁶ - ⁶C₁ x⁵ + ⁶C₂ x⁴ - ⁶C₃ x³ + ⁶C₄ x² - ⁶C₅ x + ⁶C₆)]

= 2 [⁶C₀ x⁶ + ⁶C₂ x⁴ + ⁶C₄ x² + ⁶C₆]

= 2 [x⁶ + 15x⁴ + 15x² + 1]

Putting x = √2

(x + 1)⁶ + (x - 1)⁶ = 2 [(√2)⁶ + 15 (√2)⁴ + 15 (√2)² + 1]

= 2 [8 + 15 x 4 + 15 x 2 + 1]

= 2 [8 + 60 + 30 + 1]

= 2 x 99

= 198

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NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.1 Question 12

Find (x + 1)⁶ + (x - 1)⁶. Hence or otherwise evaluate (√2 + 1)⁶ + (√2 - 1)⁶

Summary:

Using binomial theorem, (x + 1)⁶ + (x - 1)⁶ =2 [x⁶ + 15x⁴ + 15x² + 1]. Hence (√2 + 1)⁶ + (√2 - 1)⁶ = 198