The cubic equation left| begin{array}{ccc} 0 | vedclass

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(B) Let the determinant be $D$. Expanding the determinant along the first row:$D = 0 - (a-x)[(-a-x)(0) - (c-x)(-b-x)] + (b-x)[(-a-x)(-c-x) - 0(-b-x)]$

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$ac = ab + bc$

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(B) Let the determinant be $D$. Expanding the determinant along the first row:$D = 0 - (a-x)[(-a-x)(0) - (c-x)(-b-x)] + (b-x)[(-a-x)(-c-x) - 0(-b-x)]$$D = -(a-x)[-(c-x)(-b-x)] + (b-x)[(-a-x)(-c-x)]$$D = (a-x)(c-x)(-b-x) + (b-x)(a+x)(c+x)$Expanding these terms:$(a-x)(c-x)(-b-x) = (ac - ax - cx + x^2)(-b-x) = -abc - acx + abx + ax^2 + bcx + cx^2 - bx^2 - x^3 = -x^3 + (a-b+c)x^2 + (ab+bc-ac)x - abc$$(b-x)(a+x)(c+x) = (b-x)(ac + ax + cx + x^2) = abc + abx + bcx + bx^2 - acx - ax^2 - cx^2 - x^3 = -x^3 + (b-a-c)x^2 + (ab+bc-ac)x + abc$Adding them together:$D = -2x^3 + 2(ab+bc-ac)x = 0$$-2x(x^2 - (ab+bc-ac)) = 0$For the equation to have a repeated root,the discriminant of the quadratic part $x^2 - (ab+bc-ac) = 0$ must be zero,or the root $x=0$ must be a repeated root.If $x=0$ is a repeated root,then the constant term must be zero,which is already true. However,for $x=0$ to be a repeated root,the coefficient of $x$ must also be zero:$ab+bc-ac = 0 \Rightarrow ac = ab+bc$.

The cubic equation $\left| \begin{array}{ccc} 0 & a-x & b-x \\ -a-x & 0 & c-x \\ -b-x & -c-x & 0 \end{array} \right| = 0$ has a repeated root in $x$. Then:

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