solving the cubic (1 Viewer)

untouchablecuz · Mar 8, 2010

$\\ax^3+bx+c=0\\$Let $x=p+q\\\left[a(p^3+q^3) \right ]+[3pqa](p+q)=[-c]+[-b](p+q)\\$Equating terms and rearranging, $\\ p^3+q^3=-\frac{c}{a} $ and $p^3q^3=-\frac{b^3}{27a^3}\\p^3 $ and $q^3 $ are solutions to the quadratic equation $\omega ^2+\frac{c}{a}\omega-\frac{b^3}{27a^3}=0\\$Finding $p^3 $ and $q^3 $ by the quadratic formula, we can solve the cubic.$

my query is with the step that involves equating terms in (p+q) and then equating everything else. my lecturer didn't justify this step (just told us to accept it), but it seems to workout.

why can we do this?

Trebla · Mar 9, 2010

I think it's more of finding a specific case as opposed to the most general case. I don't think you can do such a thing in general unless you're equating coefficients.

solving the cubic (1 Viewer)

untouchablecuz

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Trebla

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