Question

y

⑶の解き方を教えてください。

次の式を展開せよ.
(1) (a-2)³
(3)
(2)(x-
(x−y+2)(x² + xy + y² −2x+2y+4)
解答
8
(1) a³-6a² + 1297

Answer 1

きらうる

＝{(x-y)+2}{(x²+xy+y²)-2(x-y)+4}
分配法則で展開
＝(x-y)(x²+xy+y²)-2(x-y)²+4(x-y)
　+2(x²+xy+y²)-4(x-y)+8
＝(x³-y³)-2(x²-2xy+y²)+2(x²+xy+y²)+8
＝x³-y³+6xy+8