The coefficient of x² in the expansion of the product (2−x²)((1+2x+3x²)⁶+(1−x²)⁶) is

Let a = ((1+2x+3x²)⁶+(1−x²)⁶)
Therefore, Coefficient of x² in the expansion of the product (2−x²)((1+2x+3x²)⁶+(1−x²)⁶) = 2(Coefficient of x² in a) − (Constant term in a)
In the expansion of ((1+2x+3x²)⁶+(1−x²)⁶),
Constant = 1+1 = 2
Coefficient of x² = Coefficient of x² in (₆C₀(1+2x)⁶(3x²)⁰) + Coefficient of x² in (₆C₁(1+2x)⁵(3x²)) + ₆C₂(1)²(3x²)²
= ₆C₂(4) + 6 × 3 = 60 + 18 = 78
Then, coefficient of x² in (2−x²)((1+2x+3x²)⁶+(1−x²)⁶) = 2 × 78 − 2 = 156 − 2 = 154
Let's reconsider the calculation:
Let a = ((1+2x+3x²)⁶+(1−x²)⁶)
Coefficient of x² in the expansion of the product (2−x²)((1+2x+3x²)⁶+(1−x²)⁶) = 2(Coefficient of x² in a) − (Constant term in a)
In the expansion of ((1+2x+3x²)⁶+(1−x²)⁶),
Constant term = 1 + 1 = 2
Coefficient of x²:
In (1+2x+3x²)⁶, the coefficient of x² is obtained from:
(1) ₆C₂(2x)² + ₆C₁ (3x²) = 15(4x²) + 6(3x²) = 60x² + 18x² = 78x²
In (1−x²)⁶, the coefficient of x² is obtained from:
(2) ₆C₁(-x²) = -6x²
Therefore, the coefficient of x² in a is 78 + (-6) = 72
So, the coefficient of x² in (2−x²)((1+2x+3x²)⁶+(1−x²)⁶) is 2(72) - 2 = 144 - 2 = 142
There must be some error in my calculations or the given options and solution. I cannot find a way to reach the provided options using standard binomial expansion. There may be an error in the question or provided solution.