If ax^2 + bx + c = 0 and bx^2 + cx + a = 0 ha | vedclass

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(B) Let $\alpha$ be the common root of the equations $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$.Then,$a\alpha^2 + b\alpha + c = 0$ and $b\alpha^2 + c

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$3$

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(B) Let $\alpha$ be the common root of the equations $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$.Then,$a\alpha^2 + b\alpha + c = 0$ and $b\alpha^2 + c\alpha + a = 0$.Using the method of cross-multiplication:$\frac{\alpha^2}{b(a) - c(c)} = \frac{\alpha}{c(b) - a(a)} = \frac{1}{a(c) - b(b)}$$\frac{\alpha^2}{ab - c^2} = \frac{\alpha}{bc - a^2} = \frac{1}{ac - b^2}$From the first and third terms,$\alpha^2 = \frac{ab - c^2}{ac - b^2}$.From the second and third terms,$\alpha = \frac{bc - a^2}{ac - b^2}$.Since $\alpha^2 = (\alpha)^2$,we have:$\frac{ab - c^2}{ac - b^2} = \left( \frac{bc - a^2}{ac - b^2} ight)^2$$(ab - c^2)(ac - b^2) = (bc - a^2)^2$$a^2bc - ab^3 - ac^3 + b^2c^2 = b^2c^2 - 2a^2bc + a^4$$a^4 + ab^3 + ac^3 = 3a^2bc$Dividing both sides by $abc$ (since $a, b, c 
eq 0$):$\frac{a^3}{bc} + \frac{b^2}{c} + \frac{c^2}{b} = 3$ (This is not the direct path,let's simplify the equation $a^4 + ab^3 + ac^3 = 3a^2bc$)Divide by $a$ (since $a 
eq 0$):$a^3 + b^3 + c^3 = 3abc$Therefore,$\frac{a^3 + b^3 + c^3}{abc} = 3$.

If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root and $a, b, c$ are non-zero real numbers,then $\frac{a^3 + b^3 + c^3}{abc} = $

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