Hi Priyansha, the question is for what values of b the given equation has three real roots.

Coming to explanation, if b=1 then the roots are x=a and i, -i which are imaginary. Hence, b is not equal 1.

Hope you got it.

Hi ,how do we decide the nature of roots.

Hi, for a cubic equation to find whether the roots are real or not we must know at least one root. Then by using the obtained quad equation we can find whether the roots are real or not.

HI Pooja,
If you take the value of b to be 1.
The equation will be of the form :
$$x^3-ax^2+x-a$$.
This can be rewritten as :
$$x^2\left(x-a\right)+\left(x-a\right)$$
$$\left(x-a\right)\left(x^2+1\right)$$.
x = a is one root and for$$\left(x^2+1\right)$$ we only have 2 complex roots.
Hope this helps