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We'll first compute these intersections by setting x=0 and y=0 consecutively. This gives y=a-1 and a/(x-1)^2-1=0. Hence we find (x-1)^2=a, so x=1+-sqrt(a). As we have a>1 and we want the intersection w...

First we note we can factorise the n^3 out of the expression, giving n^3(n^2-1). Secondly, we can see that the second term is a difference of two squares, allowing is to factorise the total to n^3(n-1)(n+...

We know that non-real roots appear in complex conjugate pairs. Hence when we know one root, we know both of them. Then, as we can factorise a quadratic in it's linear factors, we know our quadratic is a c...