求：一元三次方程的解法以及像：x^3+3x^2+3x+1=05x^3+11x^2+7x+1=0x^3+4x^2+11x+14=0x^3-7x^2-80x+26=0

Question

求：一元三次方程的解法
以及像：
x^3+3x^2+3x+1=0
5x^3+11x^2+7x+1=0
x^3+4x^2+11x+14=0
x^3-7x^2-80x+26=0

Answer 1

① x^3+3x^2+3x+1=0
（x³+1）+（3x²+3x）=0
(x+1)(x²-x+1)+3x(x+1)=0
(x+1)(x²-x+1+3x)=0
(x+1)(x²+2x+1)=0
(x+1)(x+1)²=0
∴x=-1
② 5x^3+11x^2+7x+1=0
( 5x³+5x²)+(6x²+6x)+(x+1)=0
5x²(x+1)+6x(x+1)+(x+1)=0
(x+1)(5x²+6x+1)=0
(x+1)(5x+1)(x+1)=0
∴x=-1 或x=-1/5
③ x^3+4x^2+11x+14=0
x³+4x²+4x+7x+14=0
x(x+2)²+7(x+2)=0
(x+2)(x²+2x+7)=0
x=-2,x²+2x+7=0无解
④ x^3-7x^2-80x+26=0
x³-13x²+6x²-80x+26=0
x²(x-13)+2(3x²-40x+13)=0
x²(x-13)+2(x-13)(3x-1)=0
(x-13)(x²+6x-2)=0
∴x=13,或x²+6x-2=0(在有理数范围内无解)
在实数范围内x²+6x-2=0
(x+3)²=11
x=(±根号11)-3