Question

(X^2 - 5x + 6)^3 - ( x^2 - 5x + 6) = 0

Answer 1

$x^{2} -5x+6 = y$
$(x^{2} -5x+6)^{3}-(x^{2} -5x+6) = y^{3}-y$
$y^{3}-y = 0$
$entao: y(y^{2}-1) = 0$
$portanto: y = 0 \ ou \ y^{2}-1 = 0 \\ y = 0\ temos: \ x^{2}-5x+6 = 0\\ y^{2} - 1 = 0 \ temos: \ (y-1).(y+1) = y^{2} - 1 = 0$
$---------------------\\ \ o \ conjunto \ solucao \ eh \ obtido \ resolvendo \ 3 \ equacoes:\\ y-1 = 0\\ y+1 = 0 \\ y = 0$
$y = x^{2}- 5x + 6 = 0 \\ S= \frac{-b}{a} =\frac{-(5)}{1} \\ \\ P = \frac{c}{a} =\frac{6}{1} \\\\ x = 2 \ ou \ x = 3$
$y - 1 = x^{2}-5x+ 5 = 0 \\ y + 1 = x^{2}-5x+7 = 0 \\ \beta = b^{2} - 4ac \\ \\ x = \frac{b+-\sqrt{\beta}} {2a} \\\\ x = \frac{-5+i\sqrt{3} }{2} \\ \\ x = \frac{-5-i\sqrt{3} }{2} \\ \\ x = \frac{-5-\sqrt{5} }{2} \\ \\ x = \frac{-5+\sqrt{5} }{2}$

S = {2,3,$\frac{-5+i\sqrt{3} }{2}$,$\frac{-5-i\sqrt{3} }{2}$,$\frac{-5-\sqrt{5} }{2}$,$\frac{-5+\sqrt{5} }{2}$}