此題的答案是 3abc(bz-cy)(cx-az)(ay-bx)有過(guò)程嗎，大神？首先，有這樣兩個(gè)公式：a³+b³=(a+b)(a²-ab+b²)，a³+b³=(a+b)³-3a²b-3ab²那么，可以推導(dǎo)出下面這個(gè)關(guān)系式：x³+y³+z³-3xyz =[(x+y)³-3x²y-3xy²]+z³-3xyz=[(x+y)³+z³]-(3x²y+3xy²+3xyz)=(x+y+z)[(x+y)²-(x+y)z+z²]-3xy(x+y+z) =(x+y+z)(x²+y²+2xy-xz-yz+z²)-3xy(x+y+z) =(x+y+z)(x²+y²+z²-xy-xz-yz)所以 x³+y³+z³=(x+y+z)(x²+y²+z²-xy-xz-yz)+3xyz 回到你的題目：a^3(bz-cy)^3+b^3(cx-az)^3+c^3(ay-bx)^3=(abz-acy)³+(bcx-abz)³+(acy-bcx)³把這三項(xiàng)分別對(duì)應(yīng)推導(dǎo)出的公式里的x、y、z，那么這三項(xiàng)的和正好為 (abz-acy)+(bcx-abz)+(acy-bcx)=0所以原式就等于公式里的 3xyz即三項(xiàng)之積的三倍 3abc(bz-cy)(cx-az)(ay-bx)