Form factored of the polynomial discriminant

[Bruno Tolentino]

I wrote x² - (a + b)x + (ab) in the wolfram and polynomial discriminant was: a² - 2ab + b². Factoring: (a-b)²

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So, I wrote x³ - (a+b+c) x² + (bc+ca+ab) x - (abc) and the polynomial discrimant given was:

Factoring: (b-c)² (c-a)² (a-b)²

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Now, I wrote x² - 2Ax + B² and the polynomial discriminant is: A² - B². Factoring: (A+B)(A-B)

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At last, but, no minus important, given this polynomial: x³ - 3 A x² + 3 B² x - C³, which is the factored form of the polynomial discriminant?

 

[andrewkirk]

The formula for the discriminant of a cubic is here. Plug your coefficients into that and see what happens.

I see no reason to expect that an easy factorisation is available. The three preceding examples you gave are all simple. The first two polynomials are just $(x-a)(x-b)$ and $(x-a)(x-b)(x-c)$ respectively, and the third one is only a quadratic. The fourth, about which you ask, is something else entirely.

 

[Bruno Tolentino]

The formula for the discriminant of a cubic is here. Plug your coefficients into that and see what happens.

I know how find it! The problem is the factorization...