Question

15．已知x+y+z=3，求（-x+y+z）³+（x-y+z）³+（x+y-z）³+24xyz的值．

Answer 1

分析 令$\left\{\begin{array}{l}{-x+y+z=r}\\{x-y+z=s}\\{x+y-z=t}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=\frac{s+t}{2}}\\{y=\frac{r+t}{2}}\\{z=\frac{r+s}{2}}\end{array}\right.$，代入化简即可得出．

解答 解：令$\left\{\begin{array}{l}{-x+y+z=r}\\{x-y+z=s}\\{x+y-z=t}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=\frac{s+t}{2}}\\{y=\frac{r+t}{2}}\\{z=\frac{r+s}{2}}\end{array}\right.$，
∴原式=r³+s³+t³+24×$\frac{s+t}{2}$×$\frac{r+t}{2}$×$\frac{r+s}{2}$=r³+s³+t³+3（s+t）（r+t）（r+s）=（r+s+t）³=（x+y+z）³=3³=27．

点评 本题考查了乘法公式的应用、换元法，考查了推理能力与计算能力，属于中档题．