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