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

αu + βv = 18,

γu + δv = 20,

then u + v equals:

(1) 5

(2) 4

(3) 3

(4) 8

(x + √(x³ − 1))⁵ + (x − √(x³ − 1))⁵.

1. Expansion of the Expression

Let the expression be: 
E = (x + √(x³ − 1))⁵ + (x − √(x³ − 1))⁵

This is of the form (A + B)⁵ + (A − B)⁵ where:
A = x
B = √(x³ − 1)

The identity for even binomial powers is: 
(A + B)ⁿ + (A − B)ⁿ = 2 × Σ (C(n, k) A^n−k B^k), where k is even.

Using n = 5 and even k = 0, 2, 4:

E = 2 [ C(5,0)A⁵ + C(5,2)A³B² + C(5,4)A B⁴ ]

Substitute:
B² = x³ − 1,
B⁴ = (x³ − 1)².

So, 
E = 2 [ x⁵ + 10x³(x³ − 1) + 5x(x³ − 1)² ]

Expanding the terms:

• x⁵
• 10x³(x³ − 1) = 10x⁶ − 10x³
• 5x(x³ − 1)² = 5x(x⁶ − 2x³ + 1) = 5x⁷ − 10x⁴ + 5x

Combine all: 
E = 2[ 5x⁷ + 10x⁶ + x⁵ − 10x⁴ − 10x³ + 5x ]

E = 10x⁷ + 20x⁶ + 2x⁵ − 20x⁴ − 20x³ + 10x 

2. Coefficients

From the polynomial:

• α (coeff. of x⁷) = 10
• β (coeff. of x⁵) = 2
• γ (coeff. of x³) = −20
• δ (coeff. of x) = 10

3. Solve the System of Equations

We are given:

10u + 2v = 18 
−20u + 10v = 20

Simplify

Divide the first equation by 2: 
5u + v = 9 → v = 9 − 5u

Divide the second equation by 10: 
−2u + v = 2

Substitute:

−2u + (9 − 5u) = 2 
−7u + 9 = 2 
−7u = −7 
u = 1

v = 9 − 5(1) = 4

4. Final Answer

u + v = 1 + 4 = 5

Correct answer: 5