解方程：x^3=1 x^3=-1这个方程如何求解？（复数根）
\u89e3\u4e00\u5143\u4e09\u6b21\u65b9\u7a0b;x^3+x+1=0\uff0c\u8981\u8fc7\u7a0b\u4ee4x=u+v,\u5219\u539f\u5f0f\u53d8\u4e3a\uff08u+v\uff09^3=-(u+v)-1
\u5219\uff1au^3+v^3+3uv(u+v)=-1-(u+v)
\u5de6\u53f3\u5bf9\u5e94\u76f8\u7b49\u5f97\uff1au^3+v^3=-1\uff0c3uv=-1\u3002
\u5219\uff1au^3+v^3=-1, u^3v^3=-1/27
\u6839\u636e\u97e6\u8fbe\u5b9a\u7406\uff1av^3\u548cu^3\u662fx^2+x-1/27=0\u7684\u4e24\u4e2a\u6839\u3002
\u89e3\u5f97\uff1au^3=-1/2+1/2\u4e58\u4ee5\u6839\u53f7\u4e0b31/27
v^3=-1/2-1/2\u4e58\u4ee5\u6839\u53f7\u4e0b31/27
\u6839\u636ex^3=1\u67093\u4e2a\u89e3\uff0cx1=1\uff0cx2=w\uff0cx3=w^2 , \u8fd9\u91ccw=(-1+\u6839\u53f73i)/2
x=u1+v1 \u89e3\u5f97u\u662f3\u4e2a\u89e3\uff0cu1=3\u6b21\u6839\u53f7\u4e0b-1/2+1/2\u4e58\u4ee5\u6839\u53f7\u4e0b31/27\uff0cu2=u1w,u3=u1w^2
\u540c\u7406v1=3\u6b21\u6839\u53f7\u4e0b-1/2-1/2\u4e58\u4ee5\u6839\u53f7\u4e0b31/27\uff0cv2=v1w\uff0cv3=v1w^2
\u6240\u4ee5x1=u1+v1\uff0cx2=u1w+v1w^2\uff0cx3=v1w+u1w^2
\u6269\u5c55\u8d44\u6599\uff1a
\u56e0\u5f0f\u5206\u89e3\u6cd5\u4e0d\u662f\u5bf9\u6240\u6709\u7684\u4e09\u6b21\u65b9\u7a0b\u90fd\u9002\u7528\uff0c\u53ea\u5bf9\u4e00\u4e9b\u7b80\u5355\u7684\u4e09\u6b21\u65b9\u7a0b\u9002\u7528\uff0e\u5bf9\u4e8e\u5927\u591a\u6570\u7684\u4e09\u6b21\u65b9\u7a0b\uff0c\u53ea\u6709\u5148\u6c42\u51fa\u5b83\u7684\u6839\uff0c\u624d\u80fd\u4f5c\u56e0\u5f0f\u5206\u89e3\u3002\u5f53\u7136\uff0c\u5bf9\u4e00\u4e9b\u7b80\u5355\u7684\u4e09\u6b21\u65b9\u7a0b\u80fd\u7528\u56e0\u5f0f\u5206\u89e3\u6c42\u89e3\u7684\uff0c\u5f53\u7136\u7528\u56e0\u5f0f\u5206\u89e3\u6cd5\u6c42\u89e3\u5f88\u65b9\u4fbf\uff0c\u76f4\u63a5\u628a\u4e09\u6b21\u65b9\u7a0b\u964d\u6b21\u3002
\u4f8b\u5982\uff1a\u89e3\u65b9\u7a0bx^3-x=0
\u5bf9\u5de6\u8fb9\u4f5c\u56e0\u5f0f\u5206\u89e3\uff0c\u5f97x(x+1)(x-1)=0\uff0c\u5f97\u65b9\u7a0b\u7684\u4e09\u4e2a\u6839\uff1ax1=0\uff1bx2=1\uff1bx3=-1\u3002
\u5361\u5c14\u4e39\u5224\u522b\u6cd5\uff1a
\u5f53\u0394=(q/2)^2+(p/3)^3>0\u65f6\uff0c\u65b9\u7a0b\u6709\u4e00\u4e2a\u5b9e\u6839\u548c\u4e00\u5bf9\u5171\u8f6d\u865a\u6839\uff1b
\u5f53\u0394=(q/2)^2+(p/3)^3=0\u65f6\uff0c\u65b9\u7a0b\u6709\u4e09\u4e2a\u5b9e\u6839\uff0c\u5176\u4e2d\u6709\u4e00\u4e2a\u4e24\u91cd\u6839\uff1b
\u5f53\u0394=(q/2)^2+(p/3)^3<0\u65f6\uff0c\u65b9\u7a0b\u6709\u4e09\u4e2a\u4e0d\u76f8\u7b49\u7684\u5b9e\u6839\u3002
\u5728\u6240\u5f97\u7684\u7ed3\u679c\u662f\u8fd1\u4f3c\u503c\u7684\u60c5\u51b5\u4e0b\uff0c\u5982\u679c\u628a\u8fd1\u4f3c\u503c\u4ee3\u5165\u539f\u65b9\u7a0b\uff0c\u90a3\u4e48\u539f\u65b9\u7a0b\u7684\u5de6\u8fb9\u4e0d\u4e3a\u96f6\uff0c\u6b64\u65f6\u7528\u4ee3\u5165\u6cd5\u68c0\u9a8c\u4e0d\u80fd\u5224\u65ad\u7ed3\u679c\u662f\u5426\u6b63\u786e\uff0c\u8981\u7528\u97e6\u8fbe\u5b9a\u7406\u68c0\u9a8c\u624d\u80fd\u5224\u65ad\u7ed3\u679c\u662f\u5426\u6b63\u786e\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1\u2014\u2014\u89e3\u4e00\u5143\u4e09\u6b21\u65b9\u7a0b
x^3=-1
\u89e3\uff1a
x^3+1=0
\uff08x+1\uff09\uff08x^2-x+1\uff09=0
x1=-1
x^2-x+1=0
\u516c\u5f0f\u6cd5\uff1ax=1/2\u00b1\u221a3i/2
\u2234x2=1/2+\u221a3i/2\uff0cx3==1/2-\u221a3i/2
x^3=1
解：（x-1）（x^2+x+1）=0
x1=1
x^2+x+1=0
公式法：x=-1/2±√3i/2
∴x2=-1/2+√3i/2，x3==-1/2-√3i/2
如果算进复数根的话确实是三个根
x^3=1 ->
x^3-1=0 ->
(x-1)(x^2+x+1)=0
x-1=0或x^2+x+1=0
x-1=0的时候x=1
x^2+x=1=0的时候，根据求根公式，可得x=(-1±√-3)/2=-1/2±√3/2*i
所以有三个根
绛旓細x^3=1 瑙ｏ細锛坸-1锛夛紙x^2+x+1锛=0 x1=1 x^2+x+1=0 鍏紡娉曪細x=-1/2卤鈭3i/2 鈭磝2=-1/2+鈭3i/2锛寈3==-1/2-鈭3i/2 鍙傝冭祫鏂欙細http://baike.baidu.com/view/466729.htm
绛旓細X^3=1 x^3-1=0 (x-1)(x^2+x+1)=0 x-1=0 x^2+x+1=0 x=1鎴 x=-1/2+鈭3*i/2 鎴 x=-1/2-鈭3*i/2
绛旓細x^3=1 锛坸-1锛夛紙x^2+x+1锛=0 x1=1 x^2+x+1=0 鍏紡娉曪細x=-1/2卤鈭3i/2 鈭磝2=-1/2+鈭3i/2,x3==-1/2-鈭3i/2
绛旓細x²+x+1=0寰锛歺=(-1 卤i鈭3)/2
绛旓細x^3-1=0,(x-1)(x^2+x+1)=0,x-1=0,x^2+x+1=0,鎵浠1=1,x2=-1/2+i鈭3/2=蠅,x3=-1/2-i鈭3/2=蠅^2
绛旓細1銆佽嫢鍦ㄥ疄鏁伴泦鍐呮潵瑙ｏ紝鍒x=1锛2銆佽嫢鍦ㄥ鏁伴泦鍐呮潵瑙ｏ紝鍒欐湁涓涓牴锛屽垎鍒槸x=1鎴杧=锛1/2锛(鈭3/2)i 鎴杧=锛1/2锛(鈭3/2)i銆
绛旓細X^3=1 x^3-1=0 (x-1)(x^2+x+1)=0 x-1=0 x^2+x+1=0 x=1鎴 x=-1/2+鈭3*i/2 鎴 x=-1/2-鈭3*i/2
绛旓細鍒濅腑锛氫竴涓В x=1 楂樹腑锛氫笁涓瑙₁=1銆亁₂=cox1/3蟺+isin1/3蟺銆亁₃=cox蟺2/3+isin蟺2/3
绛旓細鍥炵瓟锛歺鐨3娆℃柟=1 x=1 鍥犱负鏄3娆℃柟,鎵浠ヤ笉浼氭湁澶嶆暟瑙
绛旓細瑙ｅ緱锛歶^3=-1/2+1/2涔樹互鏍瑰彿涓31/27 v^3=-1/2-1/2涔樹互鏍瑰彿涓31/27 鏍规嵁x^3=1鏈3涓В锛寈1=1锛寈2=w锛寈3=w^2 , 杩欓噷w=(-1+鏍瑰彿3i)/2 x=u1+v1 瑙ｅ緱u鏄3涓В锛寀1=3娆℃牴鍙蜂笅-1/2+1/2涔樹互鏍瑰彿涓31/27锛寀2=u1w,u3=u1w^2 鍚岀悊v1=3娆℃牴鍙蜂笅-1/2-1/2涔樹互鏍瑰彿涓...
