Tìm x biết : ( áp dụng hàng đẳng thức ) ( x-2)^3 – (x+2)(x^2 – 2x + 4 ) + (2x-3)(3x-2) = 0

`(x – 2)^3 – (x + 2)(x^2 – 2x + 4) + (2x – 3)(3x – 2) = 0`

`<=> x^3 – 3x^2. 2 + 3x. 2^2 – 2^3 – (x. x^2 – x. 2x + 4x + 2x^2 – 2. 2x + 2. 4) + (2x. 3x – 2x. 2 – 3. 3x + 3. 2) = 0`

`<=> x^3 – 6x^2 + 12x – 8 – (x^3 – 2x^2 + 4x + 2x^2 – 4x + 8) + (6x^2 – 4x – 9x + 6) = 0`

`<=> x^3 – 6x^2 + 12x – 8 – x^3 + 2x^2 – 4x – 2x^2 + 4x – 8 + 6x^2 – 4x – 9x + 6 = 0`

`<=> (x^3 – x^3) + (-6x^2 + 2x^2 – 2x^2 + 6x^2) + (12x – 4x + 4x – 4x – 9x) + (-8 – 8 + 6) = 0`

`<=> -x – 10 = 0`

`<=> -x = 10`

`<=> x = -10`
Vậy `x = -10`