Consider the following cubic equation:
$$x^3 + ax^2 + bx + c = 0$$
where $a, b, c \in \mathbb{C}$.
Suppose $x \in \mathbb{C}$ satisfies this equation.
Define $y := x + \frac{a}{3}$, so that $x = y – \frac{a}{3}$.
We now substitute this in the original equation.
Stewart's Galois Theory 3rd edition (and also other sources apparently) says that, after substituting:
The equation becomes
$$y^3 + py + q = 0$$
where
$$
\begin{align}
p &= \frac{a^2 – 2a^3 + 3b}{3} \\
q &= \frac{2a^3 – 9ab + 27c}{27}
\end{align}
$$
I have tried to check this in detail myself. We have:
$$
\begin{aligned}
\left(y – \frac{a}{3}\right)^3 &= y^3 – ay^2 + \frac{a^2}{3}y – \frac{a^3}{27} \\
a\left(y – \frac{a}{3}\right)^2 = a\left(y^2 – \frac{2a}{3}y + \frac{a^2}{9} \right) &= ay^2 – \frac{2a^2}{3}y + \frac{a^3}{9} \\
b\left(y – \frac{a}{3}\right) &= by – \frac{ab}{3}\\
c &= c
\end{aligned}
$$
So, summing up everything, we get the equation:
$$y^3 + \left(-\frac{a^2}{3} + b\right) y + \left( \frac{2a^3}{27} – \frac{ab}{3} + c \right) = 0$$
So I'm getting $p = \frac{-a^2 + 3b}{3}$.
Where am I going wrong? I must be making some silly mistake during the algebraic manipulations.
Best Answer
Nevermind, I checked the 4th edition of the book and there the author writes $p = \frac{-a^2 + 3b}{3}$, so it must have been a mistake in the 3rd edition that has now been corrected.
I'm leaving this thread up so that other people who encounter this same issue while reading the 3rd edition can resolve it quickly.