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I want to find the number of quadratic equations which are unchanged by cubing their roots.

Let $ax^2 + bx + c $ be a quadratic whose roots are $ \alpha$ and $\beta $.
I know that quadratic whose roots are $\alpha^3$ and $\beta^3 $ is $a^3x^2 + (b^3 - 3abc)x + c^3 = 0$.
On comparing coefficients I am facing difficulties.
Pls help me !!

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If the quadratic equation is unchanged then the roots are unchanged. Hence either $\alpha=\alpha^3$ and $\beta=\beta^3$ or $\alpha=\beta^3$ and $\beta=\alpha^3$. For the first case the only solutions are $\alpha=0,\pm1$ and $\beta=0,\pm1$. For the second case we have that $\alpha=\beta^3=(\alpha^3)^3=\alpha^9$ hence the only real solutions are $\alpha=0,\pm1$ and $\beta=0,\pm1$ once again (If we consider complex solutions then there are more).

The number of quadratic equations is then given by the number of ways in which these roots can be combined uniquely which is $6$ quadratics.

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  • $\begingroup$ I appreciate your help. But the answer is 7 $\endgroup$ – Md Masood Apr 17 at 20:02
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Assuming you want $a\ne 0$, comparing the quadratic coefficients gives $a=\pm 1$. The third equation gives $c=0, \pm 1$. Then the second equation is one of $$b=b^3-3b\ (c=1),\quad b = b^3+3b\ (c=-1),\quad b=b^3\ (c=0),$$ which have real roots $b=0, \pm 2$, $b=0$, and $b=0, \pm 1$ respectively. This gives 7 quadratics with $a=1$: $$x^2+1,\ x^2+2x+1,\ x^2-2x+1,\ x^2-1,\ x^2,\ x^2+x,\ x^2-x.$$ There are 7 more with $a=-1$ (the negations of these 7).

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