Track / Stage 6 / 16 of 400 #1016 of 2000

Problem 1016

National olympiad, first round

Algebra Difficulty 6.0 Prove it

$\left[\underline{\text { Equations of Higher Degrees (Miscellaneous) }}\right]\left[\begin{array}{l}\text { Trigonometry (Miscellaneous). }\end{array}\right]$

a) Prove that when $4 p^{3}+27 q^{2}<0$ , the equation $x^{3}+p x+q=0$ can be reduced to the equation $y^{3}-3 b y^{2}-3 a y+b=0$ in the variable $y$ by the substitution $x=\alpha y+\beta$ .

b) Prove that the solutions of the equation (*) will be the numbers $y_{1}=\operatorname{tg} \frac{\varphi}{3}, y_{2}=\operatorname{tg} \frac{\varphi+2 \pi}{3}, y_{3}=\operatorname{tg} \frac{\varphi+4 \pi}{3}$ , where $\varphi$ is determined from the conditions:

$\sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}}, \cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}}$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

a) After the substitution, we obtain the equation $\alpha^{3} y^{3}+3 \alpha^{2} \beta y^{2}+\alpha\left(3 \beta^{2}+p\right) y+\beta^{3}+p \beta+q=0$ . The conditions $3 \beta^{2}+p=-3 \alpha^{2}$ and $\alpha^{2} \beta=-\beta^{3}-p \beta-q$ must be satisfied, from which we get $3 \beta^{3}+3 p \beta+3 q=3 \beta^{3}+p \beta, \quad \beta=-\frac{3 q}{2 p}, \quad \alpha^{2}=-\left(\frac{3 q}{2 p}\right)^{2}-\frac{p}{3}=-\frac{4 p^{3}+27 q^{2}}{12 p^{2}}>0$ .

b) $a y^{3}-3 b y^{2}-3 a y+b=0 \Leftrightarrow \frac{b}{a}=\frac{3 y-y^{3}}{1-3 y^{2}}$ . By the formula $\frac{\sin 3 \alpha}{\cos 3 \alpha}=\operatorname{tg} 3 \alpha=\frac{3 \operatorname{tg} \alpha-\operatorname{tg}^{3} \alpha}{1-3 \operatorname{tg}^{2} \alpha}$ , substituting any of the numbers $y_{1}, y_{2}, y_{3}$ into the equation yields a true equality.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.