Question

If $\begin{vmatrix}x -1&5x&7\\ x^{2} -1&x -1&8\\ 2x&3x&0\end{vmatrix} =ax^{3} +bx^{2} +cx+ d$, then $c$ is equal to

Answer 1

Given $ \begin{vmatrix}x -1&5x&7\\ x^{2} -1&x -1&8\\ 2x&3x&0\end{vmatrix} =ax^{3} +bx^{2} +cx +d$
$\Rightarrow \left(x - 1\right)\left[- 24x\right] -5x \left[- 16x\right] + 7\left[\left(x^{2} -1\right)\left(3x\right)-2x\left(x-1\right)\right]$
$= ax^{3} +bx^{2} +cx +d$
$\Rightarrow -24 x^{2} +24x +80x^{2} +7\left[3x^{3} -3x -2x^{2} +2x\right]$
$= ax^{3}+bx^{2} +cx +d$
$\Rightarrow 42x^{2} +17x +21 x^{3} = ax^{3} +bx^{2} + cx +d$
By comparing coefficients of $x$ on both sides, we get : $c = 17$