Consider the individual graphs of
and
Now there will always be one solution as they cross somewhere in the 3rd quadrant no matter what, however the values of 'a' and 'b' determine how many times they cross in the first quadrant.
As
 = x^3+a)
for
has
 > 0)
, then once this curve reaches a gradient equal to the gradient of
that is a gradient of 'b', then after that there is no possible way for the two curves to intersect. Thus we deal with this condition, and so we find when
has gradient 'b':
 =3x^2)
as we are only considering the first quadrant and so
Now if there is only 1 solution, then at
we must see
that is:
^3 + a > b\sqrt{\frac{b}{3}})
Similarly, if there are 2 real solutions, then at
we must see
that is:
^3 + a = b\sqrt{\frac{b}{3}})
Lastly, for 3 real solutions, then at
we must see
that is:
^3 + a < b\sqrt{\frac{b}{3}})
Now the equation
can be re-arranged to
and thus
and
. It follows then that
for this equation and there are 2 real solutions.
Now note that this doesn't mean there are 2 real solutions and 1 imaginary solution, it means that there are 2 DISTINCT real solutions, with one of these solutions being a double root.
Hence if we let the roots be
and
then we can say:
^2(x-\beta) = x^3-27x+54)
Expanding and equating co-efficients gives:
by equating
by equating constants.
Thus solving these simultaneously we get:
Now we know that the double root must be in the first quadrant as that is where the two curves
and
will be tangent to one another. Thus
and so we get
and
Hence
^2(x+6))
So the roots are -6 and a double root at 3.
^n(1+ \frac{1}{x})^n \equiv \frac{(1+x)^{2n}}{x^n} )
