Explanation:
Exp: A
b - a + 1
Let p(x) = x\(^{3}\) + ax\(^{2}\) + bx + c
Now, -1 is a zero of the polynomial
So, p(0) = 0
\(\Rightarrow\)(-1)\(^{3}\) + a(-1)\(^{2}\) + b(-1) + c = 0
\(\Rightarrow\)-1 + a - b + c = 0
\(\Rightarrow\)a - b + c = 1
\(\Rightarrow\)c = 1 - a + b
Now, if \(\alpha,\beta,\gamma\) are the zeroes of the cubic polynomial ax\(^{3}\) + bx\(^{2}\) + cx + d, then product of zeroes is given by
\(\alpha\beta\gamma=-\frac{\text{d}}{\text{a}}\)
So, for the given polynomial, p(x) = x\(^{3}\) + ax\(^{2}\) + bx + c
\(\alpha\beta(-1)=\frac{-\text{c}}{1}=\frac{-(1-\text{a}+\text{b})}{1}\)
\(\Rightarrow\alpha\beta=1-\text{a + b}\)
Hence, the correct answer is option (a)