Using Binomial theorem expand:
(i) (x + 1/y)11 (ii) (1 + x + x²)3
(i) We have (x + 1/y)11 = 11C₀x11 + 11C₁x10(1/y) + 11C₂x9(1/y)² + ... + 11C₁₁(1/y)11
= x11 + 11x10/y + 55x9/y² + 165x8/y³ + 330x7/y⁴ + 462x6/y⁵ + 462x5/y⁶ + 330x4/y⁷ + 165x3/y⁸ + 55x2/y⁹ + 11x/y¹⁰ + 1/y¹¹
(ii) Let y = x + x², then (1 + x + x²)3 = (1 + y)3
= 3C₀ + 3C₁y + 3C₂y² + 3C₃y³
= 1 + 3y + 3y² + y³
Substituting y = x + x²:
= 1 + 3(x + x²) + 3(x + x²)² + (x + x²)³
= 1 + 3x + 3x² + 3(x² + 2x³ + x⁴) + (x³ + 3x⁴ + 3x⁵ + x⁶)
= 1 + 3x + 6x² + 7x³ + 6x⁴ + 3x⁵ + x⁶
Hence, (1 + x + x²)3 = x⁶ + 3x⁵ + 6x⁴ + 7x³ + 6x² + 3x + 1.