Hi JeffM
a.y^3 + ( c - b^2 / 3a )y + ( d + 2.b^3 / 27.a^2 - b.c/3a) = 0
It says multiply out and simplify to get :
ay^3 + ( c - (b^2 / 3a))*y + ( d + ( 2b^3 / 27a^2 ) - bc /3a) = 0
but I cant "unravel" it!
My excuse : I'm 82 and slightly GaGa!
Pete
I am 73 and totally gaga. You say "it" says but do not quote what it says. So I am not sure whether "it" is wrong or whether you are misreading it. As far as I can remember the substitution you are that trying to apply is not relevant to a depressed cubic, which is what you seem to be working on.
\(\displaystyle a \ne 0 \text { and } ax^3 + bx^2 + cx + d = 0 \implies x^3 + x^2 * \dfrac{b}{a} + x * \dfrac{c}{a} + \dfrac{d}{a} = 0.\)
\(\displaystyle \text {Let } y = x + \dfrac{b}{3a} \implies x = y - \dfrac{b}{3a}.\)
\(\displaystyle \therefore x^3 = y^3 - 3y^2 * \dfrac{b}{3a} + 3y * \dfrac{b^2}{9a^2} - \dfrac{b^3}{27a^3} = y^3 - \dfrac{by^2}{a} + \dfrac{b^2y}{3a^2} - \dfrac{b^3}{27a^3}.\)
\(\displaystyle \text { And } x^2 * \dfrac{b}{a} = \dfrac{b}{a} \left (y^2 - 2y * \dfrac{b}{3a} + \dfrac{b^2}{9a^2} \right ) = \dfrac{by^2}{a} - \dfrac{2yb^2}{3a^2} + \dfrac{b^3}{9a^3}.\)
\(\displaystyle \text {And } x * \dfrac{c}{a} = \dfrac{c}{a} \left ( y - \dfrac{b}{3a} \right ) = \dfrac{cy}{a} - \dfrac{bc}{3a^2}.\)
\(\displaystyle \therefore x^3 + x^2 * \dfrac{b}{a} + x * \dfrac{c}{a} + \dfrac{d}{a} = \)
\(\displaystyle y^3 - \dfrac{by^2}{a} + \dfrac{by^2}{a} + y * \left ( \dfrac{b^2}{3a^2} - \dfrac{2b^2}{3a^2} + \dfrac{c}{a} \right ) - \dfrac{b^3}{27a^3} + \dfrac{b^3}{9a^3} - \dfrac{bc}{3a^2} + \dfrac{d}{a} = \)
\(\displaystyle y^3 + y * \dfrac{3ac- b^2}{3a^2} + \dfrac{2b^3 - 9abc + 27a^2d}{27a^3} = 0.\).
\(\displaystyle \text {Let } b' = \dfrac{3ac - b^2}{3a^2} \text { and } c' = \dfrac{2b^3 - 9abc + 27a^2d}{27a^3} \implies y^3 + b'y + c' = 0.\)
In other words, you can always turn a general cubic into a depressed cubic missing a term in the unknown squared.
