If the polynomial f(x) = left|begin{array}{cc | vedclass

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(A) To find the constant term of the polynomial $f(x)$, we set $x = 0$.Substituting $x = 0$ into the determinant, we get:$f(0) = \left|\begin{array}{c

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$2 - 3 \cdot 2^{b} + 2^{3b}$

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(A) To find the constant term of the polynomial $f(x)$, we set $x = 0$.Substituting $x = 0$ into the determinant, we get:$f(0) = \left|\begin{array}{ccc} (1+0)^{a} & (2+0)^{b} & 1 \ 1 & (1+0)^{a} & (2+0)^{b} \ (2+0)^{b} & 1 & (1+0)^{a} \end{array}ight| = \left|\begin{array}{ccc} 1 & 2^{b} & 1 \ 1 & 1 & 2^{b} \ 2^{b} & 1 & 1 \end{array}ight|$.Expanding the determinant along the first row:$f(0) = 1 \cdot \left|\begin{array}{cc} 1 & 2^{b} \ 1 & 1 \end{array}ight| - 2^{b} \cdot \left|\begin{array}{cc} 1 & 2^{b} \ 2^{b} & 1 \end{array}ight| + 1 \cdot \left|\begin{array}{cc} 1 & 1 \ 2^{b} & 1 \end{array}ight|$.Calculating the $2 	imes 2$ determinants:$f(0) = 1(1 - 2^{b}) - 2^{b}(1 - (2^{b})^{2}) + 1(1 - 2^{b})$.$f(0) = (1 - 2^{b}) - 2^{b}(1 - 2^{2b}) + (1 - 2^{b})$.$f(0) = 1 - 2^{b} - 2^{b} + 2^{3b} + 1 - 2^{b}$.$f(0) = 2 - 3 \cdot 2^{b} + 2^{3b}$.

If the polynomial $f(x) = \left|\begin{array}{ccc} (1+x)^{a} & (2+x)^{b} & 1 \\ 1 & (1+x)^{a} & (2+x)^{b} \\ (2+x)^{b} & 1 & (1+x)^{a} \end{array}\right|$, then the constant term of $f(x)$ is ($a$ and $b$ are positive integers).

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