Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,\,\,6{x^2}y\\
b,\,\,\,\,4{y^2}\\
c,\,\,\,\,{x^4}\\
d,\,\,\,\,{x^6} - 1\\
2,\\
a,\,\,\,8{x^3} + 36{x^2}y + 54x{y^2} + 27{y^3} = {\left( {2x + 3y} \right)^3}\\
b,\,\,\,8{x^3} + 12{x^2}y + 6x{y^2} + {y^3} = {\left( {2x + y} \right)^3}\\
c,\,\,\,{x^3} - 6{x^2}y + 12x{y^2} - 8{y^3} = {\left( {x - 2y} \right)^3}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
a,\\
{\left( {x + y} \right)^3} - {\left( {x - y} \right)^3} - 2{y^3}\\
 = \left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^3}} \right) - \left( {{x^3} - 3{x^2}y + 3x{y^2} - {y^3}} \right) - 2{y^3}\\
 = {x^3} + 3{x^2}y + 3x{y^2} + {y^3} - {x^3} + 3{x^2}y - 3x{y^2} + {y^3} - 2{y^3}\\
 = \left( {{x^3} - {x^3}} \right) + \left( {3{x^2}y + 3{x^2}y} \right) + \left( {3x{y^2} - 3x{y^2}} \right) + \left( {{y^3} + {y^3} - 2{y^3}} \right)\\
 = 6{x^2}y\\
b,\\
{\left( {x + y} \right)^2} - 2\left( {x + y} \right)\left( {x - y} \right) + {\left( {x - y} \right)^2}\\
 = {\left[ {\left( {x + y} \right) - \left( {x - y} \right)} \right]^2}\\
 = {\left( {x + y - x + y} \right)^2}\\
 = {\left( {2y} \right)^2}\\
 = 4{y^2}\\
c,\\
\left( {{x^2} - 6z} \right).\left( {{x^2} + 6z} \right) + 36{z^2}\\
 = \left[ {{{\left( {{x^2}} \right)}^2} - {{\left( {6z} \right)}^2}} \right] + 36{z^2}\\
 = \left( {{x^4} - 36{z^2}} \right) + 36{z^2}\\
 = {x^4}\\
d,\\
\left( {{x^2} + x + 1} \right).\left( {{x^2} - x + 1} \right).\left( {{x^2} - 1} \right)\\
 = \left[ {\left( {{x^2} + 1} \right) + x} \right].\left[ {\left( {{x^2} + 1} \right) - x} \right].\left( {{x^2} - 1} \right)\\
 = \left[ {{{\left( {{x^2} + 1} \right)}^2} - {x^2}} \right].\left( {{x^2} - 1} \right)\\
 = \left[ {\left( {{x^4} + 2{x^2} + 1} \right) - {x^2}} \right].\left( {{x^2} - 1} \right)\\
 = \left( {{x^2} - 1} \right).\left( {{x^4} + {x^2} + 1} \right)\\
 = \left( {{x^2} - 1} \right).\left[ {{{\left( {{x^2}} \right)}^2} + {x^2}.1 + {1^2}} \right]\\
 = {\left( {{x^2}} \right)^3} - {1^3}\\
 = {x^6} - 1\\
2,\\
a,\,\,\,8{x^3} = {\left( {2x} \right)^3};\,\,\,\,27{y^3} = {\left( {3y} \right)^3}\\
 \Rightarrow 8{x^3} + 36{x^2}y + 54x{y^2} + 27{y^3}\\
 = {\left( {2x} \right)^3} + 3.4{x^2}.3y + 3.2x.9{y^2} + {\left( {3y} \right)^3}\\
 = {\left( {2x} \right)^3} + 3.{\left( {2x} \right)^2}.3y + 3.2x.{\left( {3y} \right)^2} + {\left( {3y} \right)^3}\\
 = {\left( {2x + 3y} \right)^3}\\
b,\\
8{x^3} = {\left( {2x} \right)^3};\,\,\,\,\,\,12{x^2}y = 3.{\left( {2x} \right)^2}.y\\
 \Rightarrow 8{x^3} + 12{x^2}y + 6x{y^2} + {y^3}\\
 = {\left( {2x} \right)^3} + 3.4{x^2}.y + 3.2x.{y^2} + {y^3}\\
 = {\left( {2x} \right)^3} + 3.{\left( {2x} \right)^2}.y + 3.2x.{y^2} + {y^3}\\
 = {\left( {2x + y} \right)^3}\\
c,\\
{x^3} - 6{x^2}y + 12x{y^2} - 8{y^3}\\
 = {x^3} - 3.{x^2}.2y + 3.x.4{y^2} - 8{y^3}\\
 = {x^3} - 3.{x^2}.2y + 3.x.{\left( {2y} \right)^2} - {\left( {2y} \right)^3}\\
 = {\left( {x - 2y} \right)^3}\\
3,\\
a,\\
{a^3} + {b^3}\\
 = \left( {{a^3} + 3{a^2}b + 3a{b^3} + {b^3}} \right) - 3{a^2}b - 3a{b^2}\\
 = \left( {{a^3} + 3{a^2}b + 3a{b^3} + {b^3}} \right) - \left( {3{a^2}b + 3a{b^2}} \right)\\
 = {\left( {a + b} \right)^3} - 3ab.\left( {a + b} \right)\\
b,\\
{a^3} - {b^3}\\
 = \left( {{a^3} - 3{a^2}b + 3a{b^3} - {b^3}} \right) + 3{a^2}b - 3a{b^2}\\
 = \left( {{a^3} - 3{a^2}b + 3a{b^3} - {b^3}} \right) + \left( {3{a^2}b - 3a{b^2}} \right)\\
 = {\left( {a - b} \right)^3} + 3ab.\left( {a - b} \right)\\
4,\\
a,\\
\left( {3x + 7} \right)\left( {2x + 3} \right) - \left( {3x - 5} \right) + \left( {2x + 11} \right)\\
 = \left( {6{x^2} + 9x + 14x + 21} \right) - 3x + 5 + 2x + 11\\
 = \left( {6{x^2} + 23x + 21} \right) - 3x + 5 + 2x + 11\\
 = 6{x^2} + \left( {23x - 3x + 2x} \right) + \left( {21 + 5 + 11} \right)\\
 = 6{x^2} + 22x + 37\\
b,\\
x\left( {2x + 1} \right) - {x^2}\left( {x + 2} \right) + {x^3} - x + 3\\
 = \left( {2{x^2} + x} \right) - \left( {{x^3} + 2{x^2}} \right) + {x^3} - x + 3\\
 = 2{x^2} + x - {x^3} - 2{x^2} + {x^3} - x + 3\\
 = \left( {2{x^2} - 2{x^2}} \right) + \left( {x - x} \right) + \left( { - {x^3} + {x^3}} \right) + 3\\
 = 3\\
c,\\
{\left( {4x - 1} \right)^3} - \left( {4x - 3} \right).\left( {16{x^2} + 3} \right)\\
 = \left[ {{{\left( {4x} \right)}^3} - 3.{{\left( {4x} \right)}^2}.1 + 3.4x{{.1}^2} - {1^3}} \right] - \left( {64{x^3} + 12x - 48{x^2} - 9} \right)\\
 = \left( {64{x^3} - 48{x^2} + 12x - 1} \right) - \left( {64{x^3} - 48{x^2} + 12x - 9} \right)\\
 = 64{x^3} - 48{x^2} + 12x - 1 - 64{x^3} + 48{x^2} - 12x + 9\\
 = \left( {64{x^3} - 64{x^3}} \right) + \left( { - 48{x^2} + 48{x^2}} \right) + \left( {12x - 12x} \right) + \left( { - 1 + 9} \right)\\
 = 8
\end{array}\)

(Chỉ phần a thì giá trị của biểu thức mới phụ thuộc vào biến thôi nhé!)