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Suppose I have a cubic equation of $$x^3 + ax^2 + bx + c=0.$$ What steps would one take to eliminate the $x^2$ term? Given an elliptic curve that is not of the form $$Y^2 = X^3 + AX+B,$$ my goal would be to normalize the elliptic curve to that form with the appropriate substitutions. Handling the $Y$ side isn't a problem as all that is needed is to complete the square, but I am not sure how to get rid of the $x^2$ term on the $X$ side.

I'm not sure what subject this falls under so additional tags are welcomed.

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    $\begingroup$ Show her the morning newspaper, that should work! $\endgroup$
    – Pedro
    Commented Jul 31, 2013 at 0:11
  • $\begingroup$ @PeterTamaroff Joke not understood. :( $\endgroup$
    – mick
    Commented Jul 31, 2013 at 1:41

2 Answers 2

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Take $$x\mapsto x-\frac{a}{3}$$ to eliminate the $x^2$ term, since $${\left( {x - \frac{a}{3}} \right)^3} = {x^3} - a{x^2} + \frac{{x{a^2}}}{3} - \frac{{{a^3}}}{{27}}$$

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    $\begingroup$ I am doing an exercise related to this and will remember it forever now :). $-\dfrac a3$ is the point of inflection of the given cubic, so doing said substitution shifts the graph so that the point of inflection lies on the $y$ axis. After that, by symmetry we know $f'(x) = f'(-x)$, which means the derivative must be a parabola of the form $x^2 + \gamma$, which means that $f$ must be a depressed cubic. $\endgroup$
    – Ovi
    Commented Feb 24, 2017 at 14:48
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This is my answer for the general case of $ax^3+bx^2+cx+d$

So I'll take that $a=1$ for clean computation. I'm sure that's an easy step. Just divide the whole thing by a for any $a \ge 1$.

So we're left with $x^3+bx^2+cx+d$

So in order to get rid of the $x^2$ term you want to rewrite the thing in terms of another variable which achieves this purpose.

Let's say $y$. We rewrite in terms of $y$. Therefore $x=y+z$ for a constant $z$. How do we pick out $z$?

$(y+z)^3+b(y+z)^2+c(y+z)+d$

$y^3+z^3+3yz(y+z)+b(y^2+2yz+z^2)+cy+cz+d$

All very nice and fun to expand but I'm interested in the terms with $y^2$ only.

$3zy^2+by^2$

$(3z+b)y^2$

So if we need that term to cease to exist, then ensure that coefficient in brackets is 0.

$3z+b=0$

$z=-\dfrac{b}{3}$

There we have it. To depress a cubic you rewrite x in terms of another variable $y\text{(or whatever you want to call it)}$ where $x=y-\dfrac{b}{3}$ and b is the original $x^2$ term.

Could someone tell me what Q.E.D means so I can be using it in future answers?

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