- #1

Jhenrique

- 685

- 4

##A\lambda^2 + B##

##A\lambda^3 + B##

##A\lambda^4 + B##

##A\lambda^5 + B##

So I thought to transform the quadratic, the cubic, the quartic and the quintic function in the forms above, look:

##ax^3+bx^2+cx+d=0##

##(x=\lambda+u+v)##

##a(\lambda+u+v)^3+b(\lambda+u+v)^2+c(\lambda+u+v)+d=0##

##\\a \left ( \begin{matrix}

\lambda^3 & +3\lambda^2u & +3\lambda u^2 & +u^3\\

+3\lambda^2v & +6\lambda u v & +3u^2 v & \\

+3\lambda v^2 & +3uv^2 & & \\

+v^3 & & & \\

\end{matrix} \right )

+ b \left ( \begin{matrix}

\lambda^2 & +2\lambda u & +u^2\\

+2\lambda v & +2 u v & \\

+v^2 & &

\end{matrix} \right )

+ c (\lambda + u + v) + d = 0##

##a\lambda^3+(3a(u+v)+b)\lambda^2+(3a(u+v)^2 + 2b(u+v) + c)\lambda + (a(u+v)^3 + b(u+v)^2 + c(u+v) + d) = 0##

So, comparing the equations:

##A\lambda^3 + B = a\lambda^3+(3a(u+v)+b)\lambda^2+(3a(u+v)^2 + 2b(u+v) + c)\lambda + (a(u+v)^3 + b(u+v)^2 + c(u+v) + d) = 0##

implies that:

##\begin{cases}

A=a\\

0=(3a(u+v)+b)\\

0=(3a(u+v)^2 + 2b(u+v) + c)\\

B=(a(u+v)^3 + b(u+v)^2 + c(u+v) + d)\\

\end{cases}##

So,

##A\lambda^3 + B = 0\;\;\;\Rightarrow\;\;\;\;\lambda = \sqrt[3]{\frac{-B}{A}} \;\;\;\Rightarrow\;\;\;\; x-u-v = \sqrt[3]{\frac{-B}{A}} \;\;\;\Rightarrow\;\;\;\; x = \sqrt[3]{\frac{-B}{A}} + (u+v)\;\;\;\Rightarrow\;\;\;\; x = \sqrt[3]{\frac{-(a(u+v)^3 + b(u+v)^2 + c(u+v) + d)}{a}} + (u+v)##

Happens that

##0=3a(u+v)+b\;\;\;\Rightarrow\;\;\;\;u+v = \frac{-b}{3a}##

Conclusion:

[tex]x=\sqrt[3]{\frac{d}{a}-\frac{2b^3}{3^3a^3}-\frac{bc}{3a^2}}-\frac{b}{3a}[/tex]

This ideia is correct?