Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
VD16,\\
1,\\
{\left( {x + 1} \right)^3} = {x^3} + 3.{x^2}.1 + 3.x{.1^2} + {1^3} = {x^3} + 3{x^2} + 3x + 1\\
2,\\
{\left( {x + 3} \right)^3} = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {x^3} + 9{x^2} + 27x + 27\\
3,\\
- {\left( {3 + 2y} \right)^3} = - \left[ {{3^3} + {{3.3}^2}.2y + 3.3.{{\left( {2y} \right)}^2} + {{\left( {2y} \right)}^3}} \right]\\
= - \left( {27 + 54y + 36{y^2} + 8{y^3}} \right)\\
4,\\
{\left( {a + 2} \right)^3} = {a^3} + 3.{a^2}.2 + 3.a{.2^2} + {2^3} = {a^3} + 6{a^2} + 12a + 8\\
5,\\
{\left( {x + 3y} \right)^3} = {x^3} + 3.{x^2}.\left( {3y} \right) + 3.x.{\left( {3y} \right)^2} + {\left( {3y} \right)^3}\\
= {x^3} + 9{x^2}y + 27x{y^2} + 27{y^3}\\
VD17,\\
1,\\
{\left( {x - 2} \right)^3} = {x^3} - 3.{x^2}.2 + 3.x{.2^2} - {2^3} = {x^3} - 6{x^2} + 12x - 8\\
2,\\
{\left( {2x - 5} \right)^3} = {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.5 + 3.2x{.5^2} - {5^3}\\
= 8{x^3} - 60{x^2} + 150x - 125\\
3,\\
- {\left( {4 - 2y} \right)^3} = - \left[ {{4^3} - {{3.4}^2}.\left( {2y} \right) + 3.4.{{\left( {2y} \right)}^2} - {{\left( {2y} \right)}^3}} \right]\\
= - \left( {64 - 96y + 48{y^2} - 8{y^3}} \right)\\
= 8{y^3} - 48{y^2} + 96y - 64\\
4,\\
{\left( {2x - 3y} \right)^3} = {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.3y + 3.2x.{\left( {3y} \right)^2} - \left( {3{y^3}} \right)\\
= 8{x^3} - 36{x^2}y + 54x{y^2} - 27{y^3}\\
5,\\
{\left( {x - \frac{1}{3}y} \right)^3} = {x^3} - 3.{x^2}.\left( {\frac{1}{3}y} \right) + 3.x.{\left( {\frac{1}{3}y} \right)^2} - {\left( {\frac{1}{3}y} \right)^3}\\
= {x^3} - {x^2}y + \frac{1}{3}x{y^2} - \frac{1}{8}{y^3}\\
VD18.\\
1,\\
\left( {x + 1} \right)\left( {{x^2} - 1.x + {1^2}} \right) = {x^3} + {1^3} = {x^3} + 1\\
2,\\
\left( {y + 3} \right)\left( {{y^2} - 3y + 9} \right) = \left( {y + 3} \right).\left( {{y^2} - y.3 + {3^2}} \right)\\
= {y^3} + {3^3} = {y^3} + 27\\
3,\\
\left( {4 + x} \right).\left( {16 - 4x + {x^2}} \right) = \left( {4 + x} \right).\left( {{4^2} - 4.x + {x^2}} \right)\\
= {4^3} + {x^3} = 64 + {x^3}\\
4,\\
\left( {2x + 5} \right).\left( {4{x^2} - 10x + 25} \right) = \left( {2x + 5} \right).\left[ {{{\left( {2x} \right)}^2} - 2x.5 + {5^2}} \right]\\
= {\left( {2x} \right)^3} + {5^3} = 8{x^3} + 125\\
5,\\
\left( {2x + 3y} \right)\left( {4{x^2} - 6xy + 9{y^2}} \right) = \left( {2x + 3y} \right).\left[ {{{\left( {2x} \right)}^2} - 2x.3y + {{\left( {3y} \right)}^2}} \right]\\
= {\left( {2x} \right)^3} + {\left( {3y} \right)^3} = 8{x^3} + 27{y^3}
\end{array}\)
