How do you find all the real and complex roots of #x^3 + 3x^2 + 3x + 3 = 0#?
1 Answer
Recognise the resemblance to
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
#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 = ((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#
