Algebra question: RMO/INMO problem-solving practice

Question:

If \alpha, \beta, \gamma be the roots of the cubic equation ax^{3}+3bx^{2}+3cx+d=0. Prove that the equation in y whose roots are \frac{\beta\gamma-\alpha^{2}}{\beta+\gamma-2\alpha} + \frac{\gamma\alpha-\beta^{2}}{\gamma+\\alpha-2\beta} + \frac{\alpha\beta-\gamma^{2}}{\alpha+\beta-2\gamma} is obtained by the transformation axy+b(x+y)+c=0. Hence, form the equation with above roots.

Solution:

Given that \alpha, \beta, \gamma are the roots of the equation:

ax^{3}+3bx^{2}+3cx+d=0…call this equation I.

By relationships between roots and co-efficients, (Viete’s relations), we get

\alpha+\beta+\gamma=-\frac{3b}{a} and \alpha\beta+\beta\gamma+\gamma\alpha=\frac{3c}{a}, and \alpha\beta\gamma=-\frac{d}{a}

Now, \gamma=\frac{\beta\gamma-\alpha^{2}}{\beta+\gamma-2\alpha}=\frac{\frac{\alpha\beta\gamma}{\alpha}-\alpha^{2}}{(\alpha+\beta+\gamma)-3\alpha}=\frac{-\frac{d}{a\alpha}-\alpha^{2}}{-\frac{3b}{a}-3\alpha}=\frac{d+a\alpha^{3}}{3\alpha(b+a\alpha)}, that is,

3xy(b+ax)=d+ax^{3}, or ax^{3}-3ayx^{2}-3byx+d=0…call this equation II.

Subtracting Equation II from Equation I, we get

3(b+ay)x^{2}+3(c+by)x=0

(b+ay)x+c+by=0 since x \neq 0

axy+b(x+y)+c=0 which is the required transformation.

Now, (ay+b)x=-(by+c), that is, x=-\frac{by+c}{ay+b}

Putting this value of x in Equation I, we get

-a(\frac{by+c}{ay+b})^{3}+3b(\frac{by+c}{ay+b})^{2}-3c(\frac{by+c}{ay+b})+d=0, that is,

a(by+c)^{3}-3b(by+c)^{2}(ay+b)+3c(by+c)(ay+b)^{2}-d(ay+b)^{3}=0, which is the required equation.

Cheers,

Nalin Pithwa.

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