Question

If $1,\omega ,{{\omega }^{2}}$ are cube roots of unity find the roots of the equation ${{\left( x-1 \right)}^{3}}+8=0$

Answer 1

Hint: In this question, we first need to take 8 from the left hand side to the right hand side. Then divide both the sides with -8 and apply the cube root on both the sides. Now, equate the left hand side with each of the cube roots of unity and simplify further to get the result.

Complete step-by-step answer:
The Cube Roots of Unity
Cube roots of unity are $1,\omega ,{{\omega }^{2}}$
Where $\omega =\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}$
and ${{\omega }^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}$
Now, from the given equation in the question we have
$\Rightarrow {{\left( x-1 \right)}^{3}}+8=0$
Let us now subtract 8 on both the sides to simplify it further
$\Rightarrow {{\left( x-1 \right)}^{3}}=-8$
Let us now divide with -8 on both the sides respectively
$\Rightarrow \dfrac{{{\left( x-1 \right)}^{3}}}{-8}=1$
Now, this can be further written in the simplified form to solve further as
$\Rightarrow {{\left( \dfrac{x-1}{-2} \right)}^{3}}=1$
Let us now apply the cube root on both the sides for further simplification
$\Rightarrow {{\left( {{\left( \dfrac{x-1}{-2} \right)}^{3}} \right)}^{\dfrac{1}{3}}}=\sqrt[3]{1}$
Now, on further simplification the above equation can be further written as
$\Rightarrow \dfrac{x-1}{-2}=\sqrt[3]{1}$
As already given in the question that the cube roots of unity are $1,\omega ,{{\omega }^{2}}$
Now, on considering each of the cube roots of unity separately we get,
Let us first consider the cube root of unity as 1 and substitute in the above obtained equation
$\Rightarrow \dfrac{x-1}{-2}=1$
Now, on multiplying both sides with -2 we get,
$\Rightarrow x-1=-2$
Now, on rearranging the terms we get,
$\therefore x=-1$
Now, let us consider the cube root of unity as $\omega$ and substitute in the equation
$\Rightarrow \dfrac{x-1}{-2}=\omega$
Now, on multiplying both sides with -2 we get,
$\Rightarrow x-1=-2\omega$
Now, on rearranging the terms we get,
$\therefore x=1-2\omega$
Now, let us consider the cube root of unity as ${{\omega }^{2}}$ and substitute in the equation
$\Rightarrow \dfrac{x-1}{-2}={{\omega }^{2}}$
Now, on multiplying both sides with -2 we get,
$\Rightarrow x-1=-2{{\omega }^{2}}$
Now, on rearranging the terms and simplifying further we get,
$\therefore x=1-2{{\omega }^{2}}$
Hence, the roots of the equation ${{\left( x-1 \right)}^{3}}+8=0$are $-1,1-2\omega ,1-2{{\omega }^{2}}$

Note: It is important to note that after taking -8 to the other side if we apply the cube root there itself and find the value of x then we get only one root. So, we need to write the given equation in terms of the cube root of 1 and then simplify further to get all the three roots of the given cubic equation.
It is also to be noted that while doing the arithmetic operations or rearranging we should not neglect any of the terms or consider the wrong sign. Because it changes the corresponding equation and so the roots.