Lè lè cho cái lưỡi dài ra!!!!!!!
pt⇔(3√x−3√x−1)3+(3√x−3√x−1)=(x+1)+3√x+1⇔3√x−3√x−1=3√x+1
⇔3√x=3√x+1+3√x−1
⇔x=x+1+x−1+3(3√x+1+3√x−1).3√(x+1)(x−1)
⇔x+1+x−12+3(3√x+1+3√x−1).3√(x+1)(x−1)=0
Đặt 3√x+1=a,3√x−1=b (a>b)
⇒a3+b32+3ab(a+b)=0
⇔(a+b)(a2−ab+b2+6ab)=0
⇔[a+b=0a2+5ab+b2=0
Với a+b=0 dễ dàng tìm đc x=0
Với a2+5ab+b2=0
⇔[a=b(√21−5)2(1)a=b(−√21−5)2(2)
với (1) ta có : 23√x+1=3√x−1(√21−5)
⇔x=−8−(√21−5)38−(√21−5)3=−48(2√21−9)−32(3√21−14)=−√2728
với (2) ta có 23√x+1=3√x−1(−√21−5)
⇔x=−8−(−√21−5)38−(−√21−5)3=√2728
Vậy S={0,±√2728}