1. ## cubic discriminant

Hi guys, need help with this proof, have spent a while on it and getting no where.
I followed the hint but couldn't do anything useful

Suppose that the cubic polynomial x^3 + Ax + B factors into (x-a)(x-b)(x-c)
Prove that 4A^3 + 27B^2 = 0 if and only if two (or more) of a, b, c are the same. (Hint. Multiply out the right-hand side and compare coefficients to relate A and B to a,b,c)

Cheers for the help!

2. ## Re: cubic discriminant

Originally Posted by liedora
Hi guys, need help with this proof, have spent a while on it and getting no where.
I followed the hint but couldn't do anything useful

Suppose that the cubic polynomial x^3 + Ax + B factors into (x-a)(x-b)(x-c)
Prove that 4A^3 + 27B^2 = 0 if and only if two (or more) of a, b, c are the same. (Hint. Multiply out the right-hand side and compare coefficients to relate A and B to a,b,c)

Cheers for the help!
$\displaystyle (x-a)(x-b)(x-c)=(x^2-bx-ax+ab)(x-c)=x^3-bx^2-ax^2+acx-cx^2+cbx+acx-abc$

$\displaystyle x^3+Ax+B=x^3-bx^2-ax^2+acx-cx^2+cbx+acx-abc$

Hence,

$\displaystyle 0=(a+b+c)$

$\displaystyle A=ab+bc+ac$

$\displaystyle B=-abc$

Now, what happens if $\displaystyle a=b=c$?$\displaystyle a=b$? $\displaystyle a=c$? $\displaystyle b=c$?

3. ## Re: cubic discriminant

Hi, I got that far and then didn't know where to go, I tried substituting those values into 4A^4 + 27B^2 but I didn't know what to do from there. Or if that is even the right step to take. I must be missing something!