Hi Froco,

$$x^3=1\ =>\ x=1\ \left(\ only\ if\ x\ is\ real\right)$$

Here nothing has been mentioned about the nature of x. So you cannot take x =1

Also if you consider x = 1, the the given condition $$x^2+x+1\ =\ 0$$ wouldn't be satisfied.

Hope this helps.

I understood this method, is any other way possible to do this question?

Hi Froco,

This is only method to approach this problem.

But sir himself in the video said there are multiple ways to approach this problem.

Hi Prerna, 
There are other methods, but they can be a little unfamiliar to students with a non-math background. 
Another method to do this would lie along the same line, using the cubic roots of unity or omega. 
We would try to solve the quadratic equation. 
Upon solving the given equation, we would get the value of x as $$\frac{-1\pm\ \sqrt{\ 1-4}}{2}=-\frac{1}{2}\pm\ \frac{i\sqrt{\ 3}}{2}$$
These are known as the complex roots of unity. Upon cubing this value of x, we would get 1. 
Next, we can follow through with the same logic as in the  lecture, by splitting 2018 into 2016 and 2. 
Giving us $$1+x^2$$ in the end, which would be -x in the end
But as you can see, this is a pretty niche approach, and I would not recommend you delve extremely deep into complex and imaginary numbers as they are not prevalent from a CAT perspective. 
Hope this helps. 
Please feel free to ask any further questions you might have. 
Regards

Hi Sourish,
The answer must be the same irrespective of the method you follow, and the answer given is correct, and you obtain it to be -x. We suggest you post your solution in the doubt below with the process so that we can check and get back to you with your mistakes if you have committed any.
Regards.