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Let $f(x) = x^3 + ax + b \in Q[x]$ be irreducible. Show that if $\alpha$ is a root of $f$ then $y = \alpha^2$ satisfies $y(y + a)^2 = b^2$ Deduce that $\Delta^2(f) = −(4a^3 + 27b^2)$.

I have found the solutions using other methods. What I would like to know is why "Show that if $\alpha$ is a root of $f$ then $y = \alpha^2$ satisfies $y(y + a)^2 = b^2$" is given and how can solve it using derivatives of $f$ as well as generalize the solutions for other cubic polynomials.

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$2y(y+a) = 2y^2 + 2ay = 2\alpha^4 + 2a\alpha^2 = 2\alpha\cdot \alpha^3 + 2a\alpha^2 = 2\alpha(-a\alpha - b) + 2a\alpha^2 = -2b\alpha \to y(y+a) = -b\alpha \to y^2(y+a)^2 = b^2\alpha^2 = yb^2 \to y(y+a)^2 = b^2$.

Note: there is a typo in your post that I changed from $y(y+a)2 = b^2$ to $y(y+a)^2 = b^2$

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but Why it is given? How it is related to discriminant? Thanks for the correction.. – nnumberr May 27 '14 at 18:24

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