Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{a^3} - 3a + 3b - {b^3}\\
 = \left( {{a^3} - {b^3}} \right) - \left( {3a - 3b} \right)\\
 = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) - 3.\left( {a - b} \right)\\
 = \left( {a - b} \right)\left( {{a^2} + ab + {b^2} - 3} \right)\\
b,\\
{x^2} - {y^2} + 12y - 36\\
 = {x^2} - \left( {{y^2} - 12y + 36} \right)\\
 = {x^2} - {\left( {y - 6} \right)^2}\\
 = \left( {x - y + 6} \right)\left( {x + y - 6} \right)\\
3,\\
{\left( {x + 2} \right)^2} - {x^2} + 2x - 1\\
 = {\left( {x + 2} \right)^2} - \left( {{x^2} - 2x + 1} \right)\\
 = {\left( {x + 2} \right)^2} - {\left( {x - 1} \right)^2}\\
 = \left[ {\left( {x + 2} \right) - \left( {x - 1} \right)} \right].\left[ {\left( {x + 2} \right) + \left( {x - 1} \right)} \right]\\
 = 3.\left( {2x + 1} \right)\\
4,\\
\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right) - 24\\
 = \left[ {\left( {x + 2} \right)\left( {x + 5} \right)} \right].\left[ {\left( {x + 3} \right)\left( {x + 4} \right)} \right] - 24\\
 = \left( {{x^2} + 7x + 10} \right).\left( {{x^2} + 7x + 12} \right) - 24\\
 = \left[ {\left( {{x^2} + 7x + 11} \right) - 1} \right].\left[ {\left( {{x^2} + 7x + 11} \right) + 1} \right] - 24\\
 = {\left( {{x^2} + 7x + 11} \right)^2} - {1^2} - 24\\
 = {\left( {{x^2} + 7x + 11} \right)^2} - 25\\
 = {\left( {{x^2} + 7x + 11} \right)^2} - {5^2}\\
 = \left( {{x^2} + 7x + 11 - 5} \right).\left( {{x^2} + 7x + 11 + 5} \right)\\
 = \left( {{x^2} + 7x + 6} \right).\left( {{x^2} + 7x + 16} \right)\\
 = \left( {x + 1} \right)\left( {x + 6} \right).\left( {{x^2} + 7x + 16} \right)
\end{array}\) 

$a^3-3a+3b-b^3_{}$ 

$=(a-b)(a^2+ab+b^2)-3.(a-b)_{}$ 

$=(a-b)(a^2+ab+b^2-3)_{}$ 

$b)_{}$ $x^2-y^2+12y-36_{}$ 

$=x^2-(y^2-12y+36)_{}$ 

$=x^2-(y-6)^2_{}$ 

$=[ x-(y-6)].[ x+(y-6)]_{}$ 

$=(x-y+6)(x+y-6)_{}$ 

$c)_{}$ $(x+2)^2-x^2+2x-1_{}$ 

$=x^2+4x+4-x^2+2x-1_{}$ 

$=6x+3_{}$ 

$=3.(2x+1)_{}$ 

$d)_{}$ $(x+2)(x+3)(x+4)(x+5)-24_{}$ 

$=(x+2)(x+5)(x+4)(x+3)-24_{}$ 

$=(x^2+7x+10)(x^2+7x+12)-24_{}$ 

$Đặt_{}$ $y=x^2+7x+10_{}$ 

$y.(y+2)-24_{}$ 

$=y^2+2y-24_{}$ 

$=y^2+6y-4y-24_{}$ 

$=y.(y+6)-4.(y+6)_{}$ 

$=(y+6)(y-4)_{}$ 

$Thay_{}$ $(y+6)(y-4)_{}$ $vào_{}$ $lại_{}$ 

$=(x^2+7x+10+6)(x^2+7x+10-4)_{}$ 

$=(x^2+7x+16)(x^2+7x+6)_{}$