Guest, you are certainly correct. !
There is only 1 real roots and since most students only learn about real numbers you are 100% correct.
BUT
If you study maths for long enough you will learn about imaginary. numbers.
Two of these roots are complex roots and that means that they are partly imaginary.
It is all based on the idea that \(\sqrt{-1}=i\)
That is the basic element of the whole imaginary number system.
Lets see how this works,
I have said that one of the cubic roots is \(-1-\sqrt3\;i\)
so if this is true then
\(2^3=(-1-\sqrt3\;i)^3\)
\(LHS=8\\ RHS=\; (-1-\sqrt{3}\;i)(-1-\sqrt{3}\;i)(-1-\sqrt{3}\;i)\\ RHS=\; - \;\;\;\;(1+\sqrt{3}\;i)(1+\sqrt{3}\;i)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS=\; -\;\;\;\; (1+2\sqrt3\;i+3*-1)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS=\; -\;\;\;\; (1+2\sqrt3\;i-3)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS= \;-\;\;\;\; (-2+2\sqrt3\;i)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS= \;-\;\;\;\;-2 (1-\sqrt3\;i)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS= \;\;\;\;\;2\;\; (1-\sqrt3\;i)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS= \;\;\;\;\;2\;\; (1^2-(\sqrt3)^2(i^2))\\ RHS= \;\;\;\;\;2\;\; (1^2-(3)(-1))\\ RHS= \;\;\;\;\;2\;\; (1+3)\\ RHS=\;\;\;\;\;\ 8\\ RHS=LHS\\ \mbox{therefore one cubic root of 8 is }\:\: (-1-\sqrt{3}\;i) \)
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