If a, b, c are mutually unequal real numbers, then the value of \(\displaystyle\frac{\begin{vmatrix}1 & a & a^3\\1 & b & b^3\\1 & c & c^3\end{vmatrix}}{\begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{vmatrix}} =\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → a + b + c
In both determinants, the operations \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\) give the common factors (b − a) and (c − a) in rows 2 and 3; each determinant is then expanded along the first column.
Denominator:
\( \displaystyle \begin{vmatrix}1 & a & a^2\\0 & b-a & b^2-a^2\\0 & c-a & c^2-a^2\end{vmatrix} = (b-a)(c-a)\begin{vmatrix}1 & a & a^2\\0 & 1 & b+a\\0 & 1 & c+a\end{vmatrix} \)
\( = (b-a)(c-a)\left[(c+a)-(b+a)\right] = (b-a)(c-a)(c-b) \)
Numerator, using \(b^3 - a^3 = (b-a)(b^2+ab+a^2)\):
\( \displaystyle \begin{vmatrix}1 & a & a^3\\0 & b-a & b^3-a^3\\0 & c-a & c^3-a^3\end{vmatrix} = (b-a)(c-a)\begin{vmatrix}1 & a & a^3\\0 & 1 & b^2+ab+a^2\\0 & 1 & c^2+ac+a^2\end{vmatrix} \)
\( = (b-a)(c-a)\left[(c^2-b^2) + a(c-b)\right] \)
\( = (b-a)(c-a)(c-b)(a+b+c) \)
Since a, b, c are unequal, \((b-a)(c-a)(c-b) \ne 0\) and it cancels:
\( \displaystyle \frac{(b-a)(c-a)(c-b)(a+b+c)}{(b-a)(c-a)(c-b)} = a+b+c \)