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Homework Statement
Given a general cubic [tex]a_1x^3+b_1x^2+c_1x+d_1=0[/tex] has roots [tex]\alpha,\beta,\gamma[/tex]
find the polynomial [tex]a_2x^3+b_2x^2+c_2x+d_2=0[/tex] that has roots [tex]\alpha ^2,\beta ^2,\gamma ^2[/tex]
The Attempt at a Solution
[tex]\alpha ^2+\beta ^2+\gamma ^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma)[/tex]
Thus, [tex]\alpha ^2+\beta ^2+\gamma ^2=(\frac{-b_1}{a_1})^2-2(\frac{c_1}{a_1})=\frac{b_1^2-2a_1c_1}{a_1^2}[/tex]
Therefore, [tex]-\frac{b_2}{a_2}\equiv -\frac{b_1^2-2a_1c_1}{a_1^2}[/tex]
So the new polynomial is now in the form:
[tex]a_1^2x^3+(2a_1c_1-b_1^2)x^2+c_2x+d_2=0[/tex]
Also, [tex]\alpha ^2\beta ^2\gamma ^2=(\alpha\beta\gamma)^2[/tex]
Thus, [tex]\alpha ^2\beta ^2\gamma ^2=(\frac{d_1}{a_1})^2=\frac{d_1^2}{a_1^2}[/tex]
Therefore, [tex]-\frac{d_2}{a_2}\equiv -\frac{d_1^2}{a_1^2}[/tex]
So now the polynomial is:
[tex]a_1^2x^3+(2a_1c_1-b_1^2)x^2+c_2x+d_1^2=0[/tex]
In order to find [tex]c_2[/tex] in terms of the coefficients of the first polynomial, I'll need to express
[tex]\alpha ^2\beta ^2+\alpha ^2\gamma ^2+\beta ^2\gamma ^2[/tex] in terms of sum of roots one, two and three at a time, using the similar idea as was done to find the sum of the squared roots one at a time. However, I'm unsure how to do this. Please help.