Find the equation whose root are square of the root are $x^{3} - 2 x^{2} - x + 5 = 0$

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Solution

Let the roots of $x^{3} - 2 x^{2} - x + 5 = 0$ be $a, b, c$
$a + b + c = 2, a b + b c + c a = - 1, a b c = - 5$
$a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (a b + b c + c a) = 4 + 2 = 6$
$a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = (a b + b c + c a)^{2} - 2 a b c (a + b + c) = (- 1)^{2} - 2 (- 5) (2) = 21$
The equation whose roots are $a^{2}, b^{2}, c^{2}$ is $x^{3} - (a^{2} + b^{2} + c^{2}) x^{2} + (a^{2} b^{2} + b^{2} c^{2} + c^{2} + a^{2}) x - a^{2} b^{2} c^{2} = 0$
$⟹ x^{3} - 6 x^{2} + 21 x - 25 = 0$

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