Question

If the roots of the equation $(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0$ are real and equal, show that either $a=0\mathrm{or}(a^3+b^3+c^3)=3abc.$

Answer 1

$$\begin{gathered} \text{Given:} \\ (c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0 \\ \text{Here,} \\ a=(c^2-ab),b=-2(a^2-bc),c=(b^2-ac) \\ \text{It is given that the roots of the equation are real and equal; therefore, we have:} \\ D=0 \\ \Rightarrow (b^2-4ac)=0 \\ \Rightarrow {\{-2(a^2-bc)\}}^2-4\times (c^2-ab)\times (b^2-ac)=0 \\ \Rightarrow 4(a^4-2a^{2}bc+b^2{c}^2)-4(b^2{c}^2-a{c}^3-a{b}^3+a^{2}bc)=0 \\ \Rightarrow a^4-2a^{2}bc+b^2{c}^2-b^2{c}^2+a{c}^3+a{b}^3-a^{2}bc=0 \\ \Rightarrow a^4-3a^{2}bc+a{c}^3+a{b}^3=0 \\ \Rightarrow a(a^3-3abc+c^3+b^3)=0 \\ \mathrm{Now}, \\ a=0\text{or}{a}^3-3abc+c^3+b^3=0 \\ a=0{\text{or a}}^3+b^3+c^3=3abc \end{gathered}$$