Ta có
$(4x-1)^3 + (3-4x)(9 + 12x + 16x^2) = (8x-1)(8x+1) - (3x-5)$
$<-> (4x)^3 - 1^3 - 3.4x(4x-1) - (4x-3)[(4x)^2 + 3.4x + 3^2] = (8x)^2 - 1^2 - 3x + 5$
$<-> (4x)^3 - 1 - 12x(4x-1) - [(4x)^3 - 3^3] = 64x^2 - 3x + 4$
$<-> -48x^2 +12x +26 = 64x^2 - 3x + 4$
$<-> 112x^2 -15x - 22 = 0$