a) Consider the graph of on the cartesian plane, for real .

(Note that we require for a turning point to exist - otherwise has exactly one real root)

We will prove that both values of are either all positive or all negative.

Expanding gives

(equality occurs only when .)

Setting gives for .

Setting gives for .

Since , both possible values of are positive for positive , and negative for negative .

Suppose . Then the intercept is negative, and the gradient at that point is positive. Because all turning points of are for positive , is negative definite for and thus must have only positive roots.

Now suppose . Then the intercept is positive, and the gradient at that point is positive. Because all turning points of are for negative , is positive definite for and thus must have only negative roots.

b) Expanding gives the required answer.

c) Expanding the RHS and simplifying gives the required answer.

d) From b),

Adding gives

Applying this to c) gives

Then by replacing each variable with its cube root, we get the desired identity.

e) Let the three (real) roots be .

Now, taking the sum of roots two at a time gives .

From a) we know that is required.

From d), since the roots are real we have another requirement:

.

(Note that because the roots are either all positive or all negative from a), multiplying any two together always gives a positive number, so this inequality can be used.)

Simplifying gives .

If all three roots are real then both and are required, i.e. .

It follows that if , then not all of three roots could be real (the contrapositive).

Because has real coefficients, the complex conjugate root theorem applies - any non-real roots must come in pairs. And by the fundamental theorem of algebra, has exactly three roots. Therefore if not all of the three roots of are real, then it has exactly one real root.

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