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(A) The given determinant can be expressed as the product of two matrices:$\Delta = \left| \begin{array}{ccc} a^2 & -2a & 1 \ b^2 & -2b & 1 \ c^2 &

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(A) The given determinant can be expressed as the product of two matrices:$\Delta = \left| \begin{array}{ccc} a^2 & -2a & 1 \ b^2 & -2b & 1 \ c^2 & -2c & 1 \end{array} ight| 	imes \left| \begin{array}{ccc} 1 & 1 & 1 \ x & y & z \ x^2 & y^2 & z^2 \end{array} ight|$$= -2(a-b)(b-c)(c-a) 	imes (x-y)(y-z)(z-x) = \frac{-351}{8}$.For the cubic equation $8t^3 - 62t^2 + 43t - 7 = 0$,the roots are $t_1, t_2, t_3$. The product $(x-y)(y-z)(z-x)$ is the square root of the discriminant of the cubic equation.The discriminant $D$ of $At^3 + Bt^2 + Ct + D = 0$ is $B^2C^2 - 4AC^3 - 4B^3D - 27A^2D^2 + 18ABCD$.For $8t^3 - 62t^2 + 43t - 7 = 0$,$A=8, B=-62, C=43, D=-7$.Calculating the discriminant,we find $(x-y)^2(y-z)^2(z-x)^2 = \frac{351^2}{8^2 	imes 4} = \frac{351^2}{256}$.Thus,$|(x-y)(y-z)(z-x)| = \frac{351}{16}$.Substituting this into the determinant equation: $2 |(a-b)(b-c)(c-a)| 	imes \frac{351}{16} = \frac{351}{8}$.$|(a-b)(b-c)(c-a)| = \frac{351}{8} 	imes \frac{16}{351} 	imes \frac{1}{2} = 1$.

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