(4x - 3)^3 + (3x - 2)^3 = (7x - 5)^3
⇔ (4x - 3 + 3x - 2).[(4x-3)^2-(4x-3)(3x-2)+(3x-2)^2] = (7x - 5)^3
⇔ (7x-5).[16x^2-24x+9-(12x^2-8x-9x+6)+9x^2-12x+4) = (7x - 5)^3
⇔ (7x-5).(16x^2-24x+9- 12x^2 + 17x - 6 +9x^2-12x+4) = (7x - 5)^3
⇔ (7x - 5).(13x^2-19x+7) - (7x - 5)^3 = 0
⇔ (7x - 5).[ 13x^2-19x+7 - (7x - 5)^2 ] = 0
⇔ (7x - 5).( 13x^2-19x+7 - 49x^2 - 70x +25 ) = 0
⇔ (7x - 5).(-36x^2 + 51x - 18 ) = 0
⇔ (7x - 5).[ 3(12x^2 - 17x + 6 )] = 0
⇔ (7x - 5).[ 3(12x^2 - 8x - 9x + 6 )] = 0
⇔ (7x-5).{ 3[4x(3x-2)-3(3x-2)]} = 0
⇔ (7x-5).{ 3[(3x-2)(4x-3)]} = 0
⇔ (7x-5).(9x-6).(4x-3) = 0
⇔ 7x-5 = 0 hoặc 9x-6=0 hoặc 4x-3=0
⇒ x = $\frac{5}{7}$ hoặc x= $\frac{2}{3}$ hoặc x = $\frac{3}{4}$
Vậy x = $\frac{5}{7}$ hoặc x= $\frac{2}{3}$ hoặc x = $\frac{3}{4}$
