The equations ax²+bx+c=0 and bx²+cx+a=0, where b² − 4ac≠0 have a common root, then a^3+b^3+c^3 is to (a≠0) (1) 3abc (2) abc (3) 0 (4) 1

$$\begin{gathered} Let d be the common root \\ a{d}^2+bd+c=0 \\ b{d}^2+cd+a=0 \\ \frac{d^2}{ab-c^2}=\frac{d}{cb-a^2}=\frac{1}{ac-b^2} \\ \frac{ab-c^2}{cb-a^2}=\frac{cb-a^2}{ac-b^2} \\ \cancel{c^2{b}^2}+a^4-2a^{2}cb \\ =a^{2}cb-a{c}^3+b^2{c}^2-b^{3}a \\ {a}^4+a{c}^3+b^{3}a=3a^{2}cb \\ {\mathit{a}}^{\mathbf{3}}\mathbf{+}{\mathit{b}}^{\mathbf{3}}\mathbf{+}{\mathit{c}}^{\mathbf{3}}\mathbf{=}\mathbf{3}\mathit{a}\mathit{b}\mathit{c} \end{gathered}$$