Question

怎么求x的三次方

Answer 1

x^3
+px
+
q
=
0
的通解是：
x1
=
(
-q/2
+
(
(q/2)^2
+
(p/3)^3
)^(1/2)
)^(1/3)
+
(
-q/2
-
(
(q/2)^2
+
(p/3)^3
)^(1/2)
)^(1/3)
;
x2
=
m
*
(
-q/2
+
(
(q/2)^2
+
(p/3)^3
)^(1/2)
)^(1/3)
+
m^2
*
(
-q/2
-
(
(q/2)^2
+
(p/3)^3
)^(1/2)
)^(1/3)
;
x3
=
m^2
*
(
-q/2
+
(
(q/2)^2
+
(p/3)^3
)^(1/2)
)^(1/3)
+
m
*
(
-q/2
-
(
(q/2)^2
+
(p/3)^3
)^(1/2)
)^(1/3)
;
其中：m
=
(
-1
+
i
*
3^(1/2)
)/2
,
m^2
=
(
-1
-
i
*
3^(1/2)
)/2
而三次方程的一般形式：
ax^3
+
bx^2
+
cx
+
d
=
0
两边除以a,后设
x
=
y
-
b/3a,就可以化成
y^3
+
py
+
q
=
0
的形式：
p
=
c/a
-
b^2/(3*a^2),
q
=
(2*b^3)/(27*a^3)
-
(c*b)/(3*a^2)
+
d/a
;
然后再用上面的公式就行了。其中x1肯定是实根。