Theo vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=7\\x_1x_2=\dfrac{c}{a}=12\end{matrix}\right.\)
\(A=\dfrac{456-\left|x_1-x_2\right|}{x_1^3+x_2^3}\)
\(=\dfrac{456-\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}}{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}\)
\(=\dfrac{456-\sqrt{7^2-4\cdot12}}{7^3-3\cdot7\cdot12}=\dfrac{456-1}{343-252}=\dfrac{455}{91}=5\)