lf the roots of the equation $(\mathrm{a}^{2}-\mathrm{bc})$ $\mathrm{x}^{2}+2(\mathrm{b}^{2}-\mathrm{c}\mathrm{a})\mathrm{x}+(\mathrm{c}^{2}-\mathrm{a}\mathrm{b})=0$ are equal, then the condition is
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$ \mathrm{a}+\mathrm{b}+\mathrm{c}=0$
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$ \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3} -3abc=0$
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$ \mathrm{a}=0$ or $\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}-3\mathrm{a}\mathrm{b}\mathrm{c}=0$
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$ \mathrm{b}=0$ or $\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3} -3abc=0$
D
Correct answer
Explanation
For equal roots, the discriminant D = 0. Here, D = [2(b^2-ca)]^2 - 4(a^2-bc)(c^2-ab) = 0. Expanding this leads to 4(b^4 + c^2a^2 - 2b^2ca) - 4(a^2c^2 - a^3b - bc^3 + ab^2c) = 0. Simplifying results in b^4 + a^3b + bc^3 - 3ab^2c = 0, which factors to b(b^3 + a^3 + c^3 - 3abc) = 0.