Question

If $\left|\begin{array}{ccc}x-1& 5x& 7\\ x^2-1& x-1& 8\\ 2x& 3x& 0\end{array}\right|=a{x}^3+b{x}^2+cx+d$, then $c$ is equal to

Answer 1

Answer: 17

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