Find the value of (a² + √(a² - 1) )⁴ + (a² - √(a² - 1) )⁴

Solution:

Using binomial theorem, we will evaluate the expressions (a + b)⁴ and (a - b)⁴.

• (a + b)⁴ = ⁴C₀ a⁴ + ⁴C₁ a³b + ⁴C₂ a²b² + ⁴C₃ ab³ + ⁴C₄ b⁴
   
• (a - b)⁴ = ⁴C₀ a⁴ - ⁴C₁ a³b + ⁴C₂ a²b² - ⁴C₃ ab³ + ⁴C₄ b⁴

Therefore,

(a + b)⁴ + (a - b)⁴ = [(⁴C₀ a⁴ + ⁴C₁ a³b + ⁴C₂ a²b² + ⁴C₃ ab³ + ⁴C₄ b⁴) + (⁴C₀ a⁴

- ⁴C₁ a³b + ⁴C₂ a²b² - ⁴C₃ ab³ + ⁴C₄ b⁴)]

= 2( ⁴C₀ a⁴ + ⁴C₂ a²b² + ⁴C₄ b⁴)

= 2(a⁴ + 6a²b² + b⁴)   (Using nCr formula)

= 2a⁴ + 12a²b² + 2b⁴

Now, we will replace'a' with a² and 'b' with √(a² - 1), then we get,

(a² + √(a² - 1) )⁴ + (a² - √(a² - 1) )⁴

= 2 (a²)⁴ + 12 (a²)² (√(a² - 1))² + 2(√(a² - 1))⁴

= 2a⁸ + 12a⁴ (a² - 1) + 2 (a² - 1)²

= 2a⁸ + 12a⁶ - 12a⁴ + 2(a⁴ -2a² + 1)

= 2a⁸ + 12a⁶ - 12a⁴ + 2a⁴ - 4a² + 2

= 2a⁸ + 12a⁶ - 10a⁴ - 4a² + 2

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NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 6

Find the value of (a² + √(a² - 1) )⁴ + (a² - √(a² - 1) )⁴

Summary:

We evaluated the value of (a² + √(a² - 1) )⁴ + (a² - √(a² - 1) )⁴ to be 2a⁸ + 12a⁶ - 10a⁴ - 4a² + 2