Let left|begin{array}{ccc}x^2+x+1 & x+1 & 2x- | vedclass

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(A) Given the identity: $\left|\begin{array}{ccc}x^2+x+1 & x+1 & 2x-3 \ 3x^2-1 & x+2 & x-1 \ x^2+5x+1 & 2x+3 & x+4\end{array}ight| = ax^4+bx^3+cx^

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$29$

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(A) Given the identity: $\left|\begin{array}{ccc}x^2+x+1 & x+1 & 2x-3 \ 3x^2-1 & x+2 & x-1 \ x^2+5x+1 & 2x+3 & x+4\end{array}ight| = ax^4+bx^3+cx^2+dx+e$. To find the value of $e$, we substitute $x=0$ on both sides of the equation. Substituting $x=0$ in the determinant: $\left|\begin{array}{ccc}0^2+0+1 & 0+1 & 2(0)-3 \ 3(0)^2-1 & 0+2 & 0-1 \ 0^2+5(0)+1 & 2(0)+3 & 0+4\end{array}ight| = a(0)^4+b(0)^3+c(0)^2+d(0)+e$. This simplifies to: $\left|\begin{array}{ccc}1 & 1 & -3 \ -1 & 2 & -1 \ 1 & 3 & 4\end{array}ight| = e$. Now, expanding the determinant along the first row: $e = 1((2)(4) - (-1)(3)) - 1((-1)(4) - (-1)(1)) + (-3)((-1)(3) - (2)(1))$. $e = 1(8 + 3) - 1(-4 + 1) - 3(-3 - 2)$. $e = 1(11) - 1(-3) - 3(-5)$. $e = 11 + 3 + 15$. $e = 29$.

Let $\left|\begin{array}{ccc}x^2+x+1 & x+1 & 2x-3 \\ 3x^2-1 & x+2 & x-1 \\ x^2+5x+1 & 2x+3 & x+4\end{array}\right| = ax^4+bx^3+cx^2+dx+e$ be an identity in $x$. If $a, b, c, d$ are known, then the value of $e$ is

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