Let α, β, γ and δ be the coefficients of x⁷, x⁵, x³ and x respectively in the expansion of
(x+√(x³-1))⁵+(x-√(x³-1))⁵, x>1 . If u and v satisfy the equations ; ; α u+β v=18, ; ; γ u+δ v=20,
then u+v equals :

Solution

Maths sir · 4 steps

Adding the two expansions kills every odd power of the root, so only whole powers of x survive.

1. Odd terms cancel. Take a = x and b = √(x³-1). In (a+b)⁵ + (a-b)⁵ the terms with odd powers of b cancel and the even ones double. So only b² = x³-1 and b⁴ = (x³-1)² appear.

   (a+b)⁵+(a-b)⁵ = 2[a⁵ + 10a³b² + 5ab⁴]

2. Expand in x. This gives 2[x⁵ + 10x³(x³-1) + 5x(x⁶-2x³+1)] = 10x⁷ + 20x⁶ + 2x⁵ - 20x⁴ - 20x³ + 10x. So α = 10, β = 2, γ = -20, δ = 10.
3. Solve the pair. 10u + 2v = 18 becomes 5u + v = 9. -20u + 10v = 20 becomes -2u + v = 2. Subtracting gives 7u = 7, so u = 1 and v = 4.
4. So. u + v = 1 + 4 = 5, which is option (c).

The idea to keep. For (a+b)ⁿ + (a-b)ⁿ, write only the even-power terms of b and double them. The square roots disappear and the expansion becomes an ordinary polynomial.

Answer from the NTA final answer key. Solution by PrepWiser.