Đáp án: $ x\in\{\dfrac72, \dfrac13, -\dfrac54\}$
Giải thích các bước giải:
Ta có:
$(4x+5)^3+(2x-7)^3-(6x-2)^3=0$
$\to (4x+5)^3-(6x-2)^3+(2x-7)^3=0$
$\to ((4x+5)-(6x-2))((4x+5)^2+(4x+5)(6x-2)+(6x-2)^2)+(2x-7)^3=0$
$\to (-2x+7)(76x^2+38x+19)+(2x-7)^3=0$
$\to -(2x-7)(76x^2+38x+19)+(2x-7)^3=0$
$\to (2x-7)(76x^2+38x+19)-(2x-7)^3=0$
$\to (2x-7)(76x^2+38x+19-(2x-7)^2)=0$
$\to (2x-7)(72x^2+66x-30)=0$
$\to (2x-7)(12x^2+11x-5)=0$
$\to (2x-7)(3x-1)(4x+5)=0$
$\to x\in\{\dfrac72, \dfrac13, -\dfrac54\}$
