Cubics with one real root 
$(1) \quad f(x) = x^3- 3ax + b$
$(4) \quad f(x) = x^3 - 3hkx - (h^3 + k^3)$

By comparing the coefficients in equations $(1)$ and $(4)$, obtain two equations that relate $h$ and $k$ to $a$ and $b$.
One of these is solved to give $k$ in terms of $a$ and $h$.
Substitute this into the other equation and set $t = h^3$ to deduce a quadratic equation satisfied by $t$.
Choose $h^3$ to be one root of this quadratic and then show that the other root must be $k^3$.
By summing the cube roots of the roots of the quadratic one finds $x = h + k$.
If equation $(1)$ has exactly one real root, show that $h^3$ and $k^3$ are distinct real numbers.
I have worked out the quadratic to be $t^2 + bt + a^3$.
Not sure how to carry this forward.

Obviously, 
$$a=hk,\\b=-h^3-k^3.$$
Then
$$k=\frac ah,\\b=-h^3-\frac{a^3}{h^3}=-t-\frac{a^3}t$$
or
$$t^2+bt+a^3=0.$$
From the Vieta formulas, the sum of the roots of this quadratic equation is $-b$, and from the definition,
$$t_1+t_2=-b=h^3+k^3.$$ So if $t_1=h^3$, then $t_2=k^3$.
Now $x=h+k$ is a root of the cubic because
$$(h+k)^3-3a(h+k)+b=h^3+3hk(h+k)+k^3-3a(h+k)+b=0$$ from the defintions of $h,k$.
The cubic equation has a single real root when it has no extrema, i.e.  the derivative doesn't cancel:
$$3x^2-3a$$ is never zero for $a<0$, so that $h^2=k^2=a$ is not possible.

A personal note.
I find this exposition rather disconcerting. If prefer this one:
To solve $x^3-3ax+b=0$, decompose in two terms, $x=h+k$. Then substituting, the equation becomes
$$h^3+3hk(h+k)+k^3-3a(h+k)+b=0.$$
If we set $a=hk$, only
$$h^3+k^3+b=0$$ remains.
Now in the form
$$h^3+k^3=-b,\\h^3k^3=a^3$$
we get a sum/product problem which is easily solved by means of a bicubic equation:
$$h^6+h^3k^3=-bh^3,\\
(h^3)^2+b(h^3)+a^3=0.$$
