If the cube roots of unity are 1, [ ,,, ^2] then the roots of the equation (x - 1)^3 + 8 =…

Question

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

Accepted Answer

$${\left( {x - 1} \right)^3} + 8 = 0$$

$$ \Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$$

$$ \Rightarrow x - 1 =  - 2\,\,\,$$ or $$\,\,\, - 2\omega \,\,\,\,$$ or $$\,\,\,\, - 2{\omega ^2}$$

or $$\,\,\,x =  - 1\,\,\,$$ or $$\,\,\,1 - 2\omega \,\,\,$$ or $$\,\,\,1 - 2{\omega ^2}.$$

Solutions · 1