Answer: \(a=-\dfrac{15}{8}\).

We start with the main method. Differentiate to obtain the tangent gradient, then use the negative reciprocal for the normal gradient where needed.

The curve is

\(y=(3x-1)^{1/3}.\)

When \(x=3\),

\(y=(9-1)^{1/3}=8^{1/3}=2.\)

Differentiate:

\(\frac{dy}{dx}=(3x-1)^{-2/3}.\)

At \(x=3\),

\(\frac{dy}{dx}=8^{-2/3}=\frac14.\)

So the tangent is

\(y-2=\frac14(x-3).\)

Find its intercepts with the axes.

When \(y=0\),

\(-2=\frac14(x-3),\)

so \(x=-5\).

When \(x=0\),

\(y-2=-\frac34,\)

so \(y=\frac54\).

Thus the intercepts are

\((-5,0)\quad\text{and}\quad\left(0,\frac54\right).\)

The midpoint of \(AB\) is

\(\left(-\frac52,\frac58\right).\)

The gradient of the tangent is \(\frac14\), so the gradient of the perpendicular bisector is \(-4\).

Hence the perpendicular bisector is

\(y-\frac58=-4\left(x+\frac52\right).\)

The point \((a,a)\) lies on this line, so

\(a-\frac58=-4\left(a+\frac52\right).\)

Therefore

\(a-\frac58=-4a-10.\)

So

\(5a=-\frac{75}{8},\)

and

\(a=-\frac{15}{8}.\)

This completes the solution and gives the required result.