using cubic equation
ax^3+bx^2+cx+d=0

Let x=y-(b/3a) ----------------------(5)

a(y-(b/3a))^3+b(y-(b/3a))^2+c(y-(b/3a))+d=0

ay^3+(c-(b^2/3a))y+(d+(2b^3/27a^2)-bc/3a)=0 -------------- (1)

y^3+ay=b

let y=s-t ---------------------(4)
3st = a
s^3-t^3 = b  ---------------------------(3)

(s-t)^3+3st (s-t)=s^3-t^3

(s^3-3s^2t+3st^2-t^3)+(3s^2t-3st^2)=s^3-t^3

(a/3t)^3-t^3=b

t^6+bt^3-a^3/27 = 0 ---------------------(2)

let u=t^3

u^2 - bu - a^3/27 = 0

Here you can solve t using the equation using quadratic formula

sub t into (3) to calculate s
sub s and t into (4) to calculate y
then sub into (5) to calculate x



for 10x^3+29x^2+27x+7=0

a=10, b=29, c=27, d=7

since x=y-(b/3a)
x=y-29/81
y=x+29/81

sub into (1)

10y^3+(27-(29^2/(27*3)))y+(7+(2*29^3/(27*10^2))-(29*27)/(3*10))=0

10y^3+(27-841/81)y+(7+48778/2700-783/30)=0

10y^3+(1346/81)y+(-698/675)=0

y^3+(673/405)y+(-349/3375)=0 ------------------------- (3)

a = 673/405
b = -349/3375

sub into(2)
t^6+bt^3-a^3/27 = 0
t^6+(-349/3375)t^3-(304821217/10935)=0

let u = t^3
u^2+(-349/3375)u-(304821217/10935)=0

後面自己計

[ 本帖最後由 nissin 於 2007-4-22 03:43 PM 編輯 ]