Ian's answer is very elegant, but in case you're looking for a more computational approach, you could use the Seifert form. Namely, if you take a Seifert surface $\Sigma$ for a knot, look at the form $\Theta\colon H_1(\Sigma)\otimes H_1(\Sigma)\to \mathbb Z$ given by $\Theta(x,y)=lk(x^+,y)$ where $x^+$ is a push-off of $x$ along a consistently chosen positive normal direction. Then one can show that the Alexander polynomial is expressible as $\det(t\Theta-\Theta^T)$. Note that for a Whitehead double, there is an obvious Seifert surface with one band being a thickening of the original knot, and one band being a small twisted dual band. In particular, the Seifert form looks something like $$\left(\begin{array}{cc}0&1\0&1\end{array}\right)$$ &1\\0&1\end{array}\right)$$ which yields a trivial Alexander polynomial, which is only well-defined in this formula up to powers of t. Or you could notice that the unknot has a Seifert surface with the same Seifert form as this, by Whitehead doubling the unknot! 1 Ian's answer is very elegant, but in case you're looking for a more computational approach, you could use the Seifert form. Namely, if you take a Seifert surface \Sigma for a knot, look at the form \Theta\colon H_1(\Sigma)\otimes H_1(\Sigma)\to \mathbb Z given by \Theta(x,y)=lk(x^+,y) where x^+ is a push-off of x along a consistently chosen positive normal direction. Then one can show that the Alexander polynomial is expressible as \det(t\Theta-\Theta^T). Note that for a Whitehead double, there is an obvious Seifert surface with one band being a thickening of the original knot, and one band being a small twisted dual band. In particular, the Seifert form looks something like$$\left(\begin{array}{cc}0&1\0&1\end{array}\right) which yields a trivial Alexander polynomial, which is only well-defined in this formula up to powers of $t$. Or you could notice that the unknot has a Seifert surface with the same Seifert form as this, by Whitehead doubling the unknot!