If \begin{vmatrix}x^2+3x&x-1&x+3\\ x+1&-2x&x-4\\ x-3&x+4&3x\end{vmatrix} = ax^4+bx^3+cx^2+dx+e, then what is the value of e?

Explanation

To find e, substitute x=0 into both sides. The determinant becomes \begin{vmatrix}0&-1&3\\ 1&0&-4\\ -3&4&0\end{vmatrix}. This is a skew-symmetric matrix of order 3, whose determinant is always 0. Therefore, e = 0.

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