http://puu.sh/88lDI.png

I'm not even sure where to begin... Help will be appreciated :)

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- Apr 14th 2014, 05:51 AMBude8Cubic equation
http://puu.sh/88lDI.png

I'm not even sure where to begin... Help will be appreciated :) - Apr 14th 2014, 06:01 AMSlipEternalRe: Cubic equation
$(x-a)^2(x-b) = x^3+px+q$

Multiply out:

$(x-a)^2(x-b) = (x^2-2ax+a^2)(x-b) = x^3-(2a+b)x^2+(2ab+a^2)x-a^2b$

Equate coefficients:

$-(2a+b) = 0$

$2ab+a^2=p$

$-a^2b = q$

This is a system of three equations in four variables. So, solve for $p$ in terms of $q$. - Apr 14th 2014, 06:08 AMJeffMRe: Cubic equation
$x^3 + px + q = (x - a)^2(x - b) = (x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b =$

$x^3 - (2a + b)x^2 + a(a + 2b)x - a^2b.$ Straight forward.

The coefficients of the powers must be equal in the right hand and left expressions.

$\therefore q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.$

Proceed - Apr 14th 2014, 11:11 AMBude8Re: Cubic equation
I'm being really stupid and going wrong somewhere.

The answer is:

http://puu.sh/88G42.png

but I'm not getting that... - Apr 14th 2014, 11:30 AMromsekRe: Cubic equation
JeffM and SlipEternal have given you 3 equations that must be satisfied

$q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.$

from the last one $b=-2a$

substitute that into the first two and you get

$q=2a^3, ~p=-3a^2$

$q^2=4a^6$

$\dfrac{q^2}{4}=a^6$

$p^3=-27a^6$

$-\dfrac{p^3}{27}=a^6$

$\dfrac{q^2}{4}=-\dfrac{p^3}{27}$

$\dfrac{q^2}{4}+\dfrac{p^3}{27}=0$

$27q^2+4p^3=0$