The coefficient of $$x^2$$ in the expansion of the product $$(2 - x^2)\left\{(1 + 2x + 3x^2)^6 + (1 - 4x^2)^6\right\}$$ is:

We have to find the coefficient of $$x^{2}$$ in

$$\left(2 - x^{2}\right)\,\Big\{(1 + 2x + 3x^{2})^{6} + (1 - 4x^{2})^{6}\Big\}.$$

For convenience let us denote

$$A(x)= (1 + 2x + 3x^{2})^{6}, \qquad B(x)= (1 - 4x^{2})^{6}.$$

Then the required product is$$(2 - x^{2})\,[A(x)+B(x)].$$

When two polynomials are multiplied, the coefficient of a specific power (here $$x^{2}$$) can be obtained by pairing terms whose powers add up to that exponent. Multiplying $$2 - x^{2}$$ with $$A(x)+B(x)$$ gives

$$\bigl(2 - x^{2}\bigr)\bigl[A(x)+B(x)\bigr] = 2\,\bigl[A(x)+B(x)\bigr] \;-\; x^{2}\,\bigl[A(x)+B(x)\bigr].$$

Therefore the coefficient of $$x^{2}$$ in the final expression is

$$2 \times \bigl[\text{coefficient of }x^{2}\text{ in }A(x)+B(x)\bigr]\;-\; 1 \times \bigl[\text{coefficient of }x^{0}\text{ in }A(x)+B(x)\bigr].$$

So we need two numbers:

1. $$C_{0} =$$ coefficient of $$x^{0}$$ (constant term) in $$A(x)+B(x).$$
2. $$C_{2} =$$ coefficient of $$x^{2}$$ in $$A(x)+B(x).$$

We shall evaluate them one by one.

Constant terms

In any expansion $$(1 + \alpha)^{n},$$ the constant term is simply $$1^{n}=1.$$

Hence
$$\text{constant term of }A(x)=1,$$ because $$A(x)=(1+2x+3x^{2})^{6}$$ gives $$1$$ when we choose the $$1$$ from each of the six factors.

Similarly,
$$\text{constant term of }B(x)=1,$$ because $$B(x)=(1-4x^{2})^{6}$$ also gives $$1$$ when we choose the $$1$$ from every factor.

Thus

$$C_{0}=1+1=2.$$

Coefficient of $$x^{2}$$ in $$A(x)=(1+2x+3x^{2})^{6}$$

To obtain $$x^{2}$$ from the multinomial expansion we can proceed term-by-term:

• Choose the linear term $$2x$$ from exactly two of the six brackets and the constant $$1$$ from the remaining four. Each such choice contributes $$\bigl(2x\bigr)^{2}(1)^{4}=4x^{2}.$$ The number of ways is $$\binom{6}{2}=15.$$ Hence the contribution is $$15\times4=60.$$
• Choose the quadratic term $$3x^{2}$$ from exactly one of the six brackets and the constant $$1$$ from the other five. Each such choice contributes $$3x^{2}.$$ The number of ways is $$\binom{6}{1}=6.$$ Hence the contribution is $$6\times3=18.$$
Adding these two contributions we get

$$\text{coefficient of }x^{2}\text{ in }A(x)=60+18=78.$$

Coefficient of $$x^{2}$$ in $$B(x)=(1-4x^{2})^{6}$$

Here every non-constant term already has an even power of $$x$$ because the middle term is $$-4x^{2}.$$ To get exactly $$x^{2}$$ we must pick the term $$-4x^{2}$$ from one factor and the constant $$1$$ from the remaining five factors. Thus

$$\text{coefficient of }x^{2}\text{ in }B(x) =\binom{6}{1}\times(-4)=6\times(-4)=-24.$$

Adding the coefficients from $$A(x)$$ and $$B(x)$$ we obtain

$$C_{2}=78+(-24)=54.$$

Putting the pieces together

Recall that the coefficient required is

$$2C_{2}-C_{0}=2\times54-2=108-2=106.$$

Hence, the correct answer is Option D.