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If the cubic equation $x^3+px+q$ has roots $\alpha , \beta , \gamma $ then we know that $\alpha + \beta + \gamma =0 $, $\alpha \beta + \alpha \gamma + \beta \gamma =p $ and $\alpha \beta \gamma =-q $.

The discriminant is $\Delta = (\alpha - \beta )^2 (\alpha - \gamma )^2 (\beta - \gamma ) ^2 $.

I know the answer should be $\Delta = -4p^3 -27q^2 $ and when I substitute expressions involving roots instead of $ p$ and $q$ in the expression for $\Delta $ and try to verify it, it basically just comes out with a big mess and I haven’t been able to verify it yet. Is there a better way to show the value of the discriminant in terms of $p$ and $q$ ?

dxiv
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Anonmath101
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  • For a speculative shortcut, the discriminant is zero iff the cubic has a multiple root. That must be a root of its derivative $,3x^2+p=0,$. Separating the cubic equation, squaring and substituting $,x^2 = -p/3,$ in $,x^2(x^2+p)^2 = (-q)^2,$ gives $,-4 p^3 = 27 q^2,$, therefore $\Delta = c,(4p^3+27q^2),$ for some constant $,c,$. Calculating the discriminant e.g. for $,x^3 - x,$ gives $,c=-1,$. $,$ Note: this hand-waves the part about $,c,$ being a constant, as opposed to a non-constant polynomial in $,p,q,$, which can be proved but requires some more work. – dxiv May 05 '22 at 21:20

1 Answers1

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If the cubic equation $x^3+px+q$ has roots $\alpha , \beta , \gamma $ then we know that $\alpha + \beta + \gamma =0 $, $\alpha \beta + \alpha \gamma + \beta \gamma =p $ and $\alpha \beta \gamma =-q $.

The discriminant is $\Delta = (\alpha - \beta )^2 (\alpha - \gamma )^2 (\beta - \gamma ) ^2 $.

I know the answer should be $\Delta = -4p^3 -27q^2$

For convenience, I will employ changes of variables.

Let $~A = \alpha\beta, ~B = \alpha + \beta \implies \gamma = -B.$

Then,
$(\alpha-\beta)^2 = (B^2 - 4A).$

$[(\alpha - \gamma )(\beta - \gamma )]^2$

$ = [(2\alpha + \beta)(\alpha + 2\beta)]^2$

$ = [2\alpha^2 + 5\alpha\beta + 2\beta^2]^2$

$ = [2B^2 + A]^2 = 4B^4 + 4B^2A + A^2.$

Putting this all together, you have that

$\Delta = (B^2 - 4A)(4B^4 + 4B^2A + A^2)$

$= ~(4B^6 + 4B^4A + B^2A^2)$
$+ ~(-16B^4A - 16B^2A^2 - 4A^3)$

$$=~ 4B^6 - 12B^4A - 15B^2A^2 - 4A^3. \tag1 $$

So, the problem reduces to determining whether (1) above matches $-4p^3 -27q^2.$

$p = A + B(-B) = A - B^2.$
$q = -AB$.

Therefore,

$-4p^3 - 27q^2$

$=~ -4(A-B^2)^3 -27(-AB)^2$

$=~-4[A^3 - 3A^2B^2 + 3AB^4 - B^6] - 27A^2B^2$

$= -4A^3 + 12A^2B^2 - 12AB^4 + 4B^6 - 27A^2B^2$

$$= -4A^3 - 15A^2B^2 - 12AB^4 + 4B^6. \tag2 $$

The expressions in (1) and (2) match.

user2661923
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