How do you find all the real and complex roots of #x^3 + 3x^2 + 3x + 3 = 0#?

1 Answer
Feb 12, 2016

Answer:

Recognise the resemblance to #(x+1)^3# to simplify the cubic and solve.

Explanation:

We use the sum of cubes identity below, which can be written in the form:

#a^3+b^3=(a+b)(a^2-ab+b^2)#

One way to solve the cubic equation is to notice that it is rather similar to the binomial expansion of #(x+1)^3# ...

#0 = x^3+3x^2+3x+3#

#= x^3+3x^2+3x+1 + 2#

#=(x+1)^3+2#

#=(x+1)^3+root(3)(2)^3#

#=((x+1)+root(3)(2))((x+1)^2-root(3)(2)(x+1)+root(3)(2)^2)#

#=(x+(1+root(3)(2)))(x^2+(2-root(3)(2))x+(1-root(3)(2)+root(3)(2)^2))#

So the Real root is #x=-1-root(3)(2)# and the Complex roots are given by the quadratic formula:

#x = ((root(3)(2)-2)+-sqrt((2-root(3)(2))^2-4(1-root(3)(2)+root(3)(2)^2)))/2#

#= ((root(3)(2)-2)+-sqrt(4-2root(3)(2)+root(3)(2)^2-4+4root(3)(2)-4root(3)(2)^2))/2#

#= ((root(3)(2)-2)+-sqrt(2root(3)(2)-3root(3)(2)^2))/2#

#= (root(3)(2)-2)/2+-sqrt(3root(3)(2)^2-2root(3)(2))/2 i#