= Solution
Use the standard closed three-strand braid $B=(\sigma_1\sigma_2^{-1})^2$, a four-crossing diagram of the <figure-eight knot>. The diagram has two positive and two negative crossings, so $w(B)=0$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-24-figure-eight-braid.png]
{title=Four-crossing closed braid for the figure-eight knot, with three starting meridians}
We calculate its <Kauffman bracket>, making the inversion symmetry explicit rather than merely asserting amphichirality.
In the <Temperley-Lieb diagram algebra> on three strands put $u=e_1$, $v=e_2$. The relations are $u^2=\delta u$, $v^2=\delta v$, $uvu=u$, $vuv=v$. The bracket image of one two-crossing block is
$$
H=(A1+A^{-1}u)(A^{-1}1+Av)=1+A^{-2}u+A^2v+uv.
$$
Closing planar diagrams gives the <bracket trace on the three-strand Temperley-Lieb algebra>:
$$
\operatorname{tr}(1)=\delta^2,\quad
\operatorname{tr}(u)=\operatorname{tr}(v)=\delta,\quad
\operatorname{tr}(uv)=\operatorname{tr}(vu)=1.
$$
These values follow because the closures have respectively three, two and one circles. Expanding and reducing $H^2$ gives
$$
\begin{aligned}
H^2={}&1+(3A^{-2}+\delta A^{-4})u+(3A^2+\delta A^4)v\\
&+[4+\delta(A^2+A^{-2})]uv+vu.
\end{aligned}
$$
Taking its trace and substituting $\delta=-A^2-A^{-2}$ yields
$$
\langle B\rangle=A^8-A^4+1-A^{-4}+A^{-8}.
$$
There is no <writhe> correction, so
$$
\boxed{V_K(t)=t^2-t+1-t^{-1}+t^{-2},\qquad V_K(t^{-1})=V_K(t).}
$$
The equality is immediate from the five displayed coefficients. The same braid model supplies the meridians used in part (d).
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