Giải thích các bước giải:
$(x-3)^3-(x-3)(x^2+3x+9)×6(x-1)^2$
$=x^3-9x^2+27x-27-(x^3-27)×6(x^2-2x+1)$
$=x^3-9x^2+27x-27-(6x^3-162)(x^2-2x+1)$
$=x^3-9x^2+27x-27-6x^3(x^2-2x+1)+162(x^2-2x+1)$
$=x^3-9x^2+27x-27-6x^5+12x^4-6x^3+162x^2-324x+162$
$=-6x^5+12x^4+(x^3-6x^3)+(-9x^2+162x^2)+(27x-324x)+(-27+162)$
$=-6x^5+12x^4-5x^3+153x^2-297x+189$
$\text{Nếu đề này:}$
$(x-3)^3-(x-3)(x^2+3x+9)+6(x-1)^2$
$=x^3-9x^2+27x-27-(x^3-27)+6(x^2-2x+1)$
$=x^3-9x^2+27x-27-(x^3-27)+6(x^2-2x+1)$
$=x^3-9x^2+27x-27-x^3+27+6x^2-12x+6$
$=(x^3-x^3)+(-9x^2+6x^2)+(27x-12x)+(-27+27+6)$
$=0-3x^2+15x+6$
$=-3x^2+15x+6$
Học tốt!!!
