Đáp án:
${\left( {x + 2y + 3z} \right)^3} - {x^3} - 8{y^3} - 27{z^3} = 3\left( {2y + 3z} \right)\left( {x + 2y} \right)\left( {x + 2z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
{\left( {x + 2y + 3z} \right)^3} - {x^3} - 8{y^3} - 27{z^3}\\
= \left[ {{{\left( {x + 2y + 3z} \right)}^3} - {x^3}} \right] - \left[ {8{y^3} + 27{z^3}} \right]\\
= \left( {2y + 3z} \right)\left[ {{{\left( {x + 2y + 3z} \right)}^2} + x\left( {x + 2y + 3z} \right) + {x^2}} \right] - \left[ {{{\left( {2y} \right)}^3} + {{\left( {3z} \right)}^3}} \right]\\
= \left( {2y + 3z} \right)\left[ {{{\left( {x + 2y + 3z} \right)}^2} + x\left( {x + 2y + 3z} \right) + {x^2}} \right] - \left( {2y + 3z} \right)\left[ {{{\left( {2y} \right)}^2} - 2y.3z + {{\left( {3z} \right)}^2}} \right]\\
= \left( {2y + 3z} \right)\left[ {{{\left( {x + 2y + 3z} \right)}^2} + x\left( {x + 2y + 3z} \right) + {x^2} - {{\left( {2y} \right)}^2} + 2y.3z - {{\left( {3z} \right)}^2}} \right]\\
= \left( {2y + 3z} \right)\left[ {3{x^2} + 6xy + 9xz + 18yz} \right]\\
= 3\left( {2y + 3z} \right)\left( {{x^2} + 2xy + 3xz + 6yz} \right)\\
= 3\left( {2y + 3z} \right)\left( {x + 2y} \right)\left( {x + 2z} \right)
\end{array}$
