$(x² - 6x + 9)³ + (1- x²)³ + (6x - 10)³ = 0$
$→ (x² - 6x + 9)³ + (1-x²)³ - (10-6x)³ = 0$
$→ (x² - 6x + 9)³ + (1-x²)³ - (x² - 6x + 9 + 1 - x²)³ = 0$
Đặt $a = x² - 6x + 9 ; b = 1 - x²$
$→ a³ + b³ - (a + b)³ = 0$
$→ a³ + b³ - a³ - b³ - 3ab(a + b) = 0$
$→ - 3ab(a+b)=0$ $→ab(a+b) =0$
$TH1:a = 0 → x² - 6x + 9 = 0$
$→ (x - 3)² =0$$→ x - 3=0$ $→ x = 3$
$TH2:b = 0 → 1 - x² = 0 → x² = 1 → x = ±1$
$TH3: a + b = 0 → 10 - 6x = 0 →6x = 10→x=\frac{5}{3}$
Vậy $S={3 ; ± 1 ; \frac{5}{3}}$
