If $a x^{2} + b x + c = 0$ and $b x^{2} + c x + a = 0$ have a common root and $a, b,$ and $c$ are nonzero numbers, then find the value of $(a^{3} + b^{3} + c^{3}) / a b c$ .

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Solution

$x + 2 x + 3 x = 6 x$
$6 x = 6 x$ (infinite solution)
$I = π / 20 \int \frac{1}{3 + 5 sin x} d x$
Question is in wrong fotmate
$a x^{2} + b x + c = 0 p / q$
$b x^{2} + c x + a = 0 p / q$
$p + q = b / a / P + R = - \frac{c}{a}$
$p q = c / a / P R = \frac{a}{b}$
$(a^{3} + b^{3} + c^{3}) / a b c$
$a x^{2} + b x + c = b^{2} + c x + a$ for $P$
$a p^{2} + b p + c = b p^{2} + c p + a$
$p = x = 1$
$(a + b + c) = 0$
$\frac{[(a^{3} + b^{3} + c^{3} - 3 a b c + 3 a b c)]}{\frac{(a + b + c)^{3} + 3 a b c}{a b c}} / a b c$
$= 3$

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