Question

How do you find three cube roots of -1?

Answer 1

Hint:The given questions asks to find three cube roots of -1. In order to do that, we will at first assume the cube root of -1 for some variable. After making this assumption we will take power 3 on both the sides of the equation. After taking power 3 on both the sides we will convert the equation into such a form that the algebraic identity of the cube root can be applied and roots can be found out.

Complete step by step answer:
Now, let \[x\] be the cube root of -1.Therefore,
\[{x^3} = - 1\]
Thus,
\[{x^3} + 1 = 0\]
\[ \Rightarrow {x^3} + {1^3} = 0\]
As \[1 = {1^3}\]
Now, according to the algebraic identity of sum of cubes i.e., \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\]
We can write,
\[{x^3} + {1^3} = (x + 1)({x^2} - x + 1) = 0\]
Here, either
\[x + 1 = 0 - - - - - (1)\]
\[({x^2} - x + 1) = 0 - - - - - (2)\]
Therefore, from (1) one of the root of -1 is, \[x = - 1\]
Now for other two roots from (2)
\[ \Rightarrow ({x^2} - x + 1) = 0 \Rightarrow {x^2} - 2 \times x \times (\dfrac{1}{2}) + {(\dfrac{1}{2})^2} - (\dfrac{1}{2}) + 1 = 0\]
\[ \Rightarrow {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} = 0\]
We know, \[\dfrac{3}{4} = \dfrac{{{{\left( {\sqrt 3 } \right)}^2}}}{{{2^2}}} = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2}\]
\[ \Rightarrow {\left( {x - \dfrac{1}{2}} \right)^2} - {\left( {\dfrac{{\sqrt 3 }}{2}i} \right)^2} = 0 - - - - - (3)\]
According to algebraic identity of difference of squares, i.e., \[({x^2} - {y^2}) = (x - y)(x + y)\]
Therefore, using the above algebraic identity on equation (3)
We get,
\[ \Rightarrow \left( {x - \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i} \right)\left( {x - \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i} \right) = 0\]
Thus, we can say from the above equation that,either \[x = \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i\] or \[x = \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i\]

Hence, the three cube root -1 are \[\left( { - 1,\dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i,\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i} \right)\].

Note: One of the important things to remember about the complex numbers is, for any number of forms \[p + qi\], where p and q are real numbers, is defined as a complex number. For example: \[3 + 2i\], \[5 + i\sqrt 6 \] are both complex numbers.For any complex number\[z = p + qi\], \[p\] is the real part denoted by Re \[z\]and \[q\] is called the imaginary part denoted by Im \[z\]. Likewise, for \[\dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i\] we can say that \[\dfrac{1}{2}\] is the real part and \[\dfrac{{\sqrt 3 }}{2}i\]is the imaginary part.