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The free group $F(a, b)$ on two-generators works with the word $w=a^3b^4a^5b^6$. This works because every automorphism either takes $w$ to a word which, after cyclic reduction, does not contain $a^{\pm 3}$, or takes $w$ to a conjugate of one of the following words. $$\begin{align*} &a^{3}b^{4}a^{5}b^{6}\\ &a^{3}b^{-4}a^{5}b^{-6}\\ &a^{-3}b^{4}a^{-5}b^{6}\\ &a^{-3}b^{-4}a^{-5}b^{-6}\\ &b^{3}a^{4}b^{5}a^{6}\\ &b^{3}a^{-4}b^{5}a^{-6}\\ &b^{-3}a^{4}b^{-5}a^{6}\\ &b^{-3}a^{-4}b^{-5}a^{-6} \end{align*}$$ Thus, to prove that this works we need to prove that $w^{-1}=b^{-6}a^{-5}b^{-4}a^{-3}$ is conjugate to one of the above words. And it clearly is not.

This answer requires some knowledge of the automorphisms of free groups. The list of words are the orbits of $w$ under the automorphisms which fix the length of the generators. The observation about the other elements in the orbits not containing $a^{\pm3}$ follows from the paper What Does a Basis of $F(a,b)$ Look Like? by Cohen, Metzler and Zimmermann, although some work is required to make it follow.

Taking the same word in the triangle groups $\langle a, b; a^i, b^j, (ab)^k\rangle$ for $i, j, k>13$ works too, and here you only have to check finitely many automorphisms and you can do this by hand. Note that the "finitely many" is because you only need to check the finitely many outer automorphisms and then think about how inner automorphisms can act on this. If you are very careful, you can make it follow from the free group case but this uses small-cancellation theory and other technical stuff.