Solving some special, simple cubic equations.

The constant term = 0

Grouping

5x3 + 10x2 +5x = 0,

5x(x2 + 2x +1) = 0,

5x(x+1)(x+1) = 0

gives: x = 0, x = - 1, x = - 1

5x3 - 10x2 +9x - 18 = 0

(5x3 - 10x2 )+ (9x - 18) = 0

5x2(x - 2 ) + 9(x - 2) = 0

(x - 2 )(5x2 + 9 ) = 0

x = 2, or 5x2 = -9, x2 = -9/5

x = + 3i/5

The sum of two cubes: 
         x3 + 8 = 0,

The difference of two cubes: 
        x3- 27 = 0

(x + 2)(x2 - 2x +4) = 0
x +2 =0 or x2 - 2x +4 = 0
x = - 2 and x= (1 + (1 - 4)) which gives

x= 1 + (-3) which equals 1 + i(3)

so the three solutions are x= - 2, 1 + i(3)

(x-3)(x2 + 3x + 9) = 0

(x-3) = 0 and (x2 + 3x + 9) = 0

x = 3 and x = - 3/2 + (9/4 - 9)

so the three solution are: 
x=3, -3/2 + (3/2)i3

 

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