Đặt a = \(\sqrt[3]{3x-2}\) => a3 = 3x - 2 <=> a3 + 2 = 3x (1)
PT <=> x3 + 2 = 3a (2)
Lấy (1) trừ (2) ta được : a3 - x3 = 3(x - a)
<=> (a - x)(a2 + ax + x2) + 3(a - x) = 0
<=> (a - x)(a2 + ax + x2 + 3) = 0
<=> a - x = 0 ( Vì a2 + ax + x2 + 3 > 0 với mọi a; x)
<=> a = x
=> \(\sqrt[3]{3x-2}=x\)
<=> 3x - 2 = x3 <=> x3 - 3x + 2 = 0 <=> (x - 1).(x2 + x - 2) = 0 <=> x= 1 hoặc x = -2
vậy...
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