Solve for x:
x^3+3 x^2+3 x+3 = 0
Subtract 2 from both sides:
x^3+3 x^2+3 x+1 = -2
Factor x^3+3 x^2+3 x+1 into a perfect cube:
(x+1)^3 = -2
Taking cube roots gives (-2)^(1/3) times the third roots of unity:
x+1 = (-2)^(1/3) or x+1 = -2^(1/3) or x+1 = -((-1)^(2/3) 2^(1/3))
Subtract 1 from both sides:
x = (-2)^(1/3)-1 or x+1 = -2^(1/3) or x+1 = -((-1)^(2/3) 2^(1/3))
Subtract 1 from both sides:
x = (-2)^(1/3)-1 or x = -1-2^(1/3) or x+1 = -((-1)^(2/3) 2^(1/3))
Subtract 1 from both sides:
Answer: |x = (-2)^(1/3)-1 or x = -1-2^(1/3) or x = -1-(-1)^(2/3) 2^(1/3)
+26396 roots of equation x^3+3x^2+3x+3
x3+3x2+3x+3=0|x3+3x2+3x+1=(x+1)3(x+1)3+2=0(x+1)3=−2x+1=3√−2x=3√−2−1(1)3√−2=3√2⋅(cos(π3)+i⋅sin(π3))|tanφ=0−2 ⇒ φ=π3√−2=3√2⋅(12+i⋅√32)3√−2=3√2⋅12+i⋅√32⋅3√2x1=3√−2−1=3√2⋅12−1+i⋅√32⋅3√2x1=−0.37003947505+i⋅1.09112363597(2)3√−2=3√2⋅(cos(π3+23π)+i⋅sin(π3+23π))3√−2=3√2⋅(cos(π)+i⋅sin(π))3√−2=3√2⋅(−1+i⋅0)3√−2=−3√2x2=3√−2−1=−3√2−1x2=−2.25992104989(3)3√−2=3√2⋅(cos(π3+2⋅23π)+i⋅sin(π3+2⋅23π))3√−2=3√2⋅(cos(53π)+i⋅sin(53π))3√−2=3√2⋅(12−i⋅√32)3√−2=3√2⋅12−i⋅√32⋅3√2x3=3√−2−1=3√2⋅12−1−i⋅√32⋅3√2x3=−0.37003947505−i⋅1.09112363597
