Let $A$ be an abelian variety of dimension g, and $x\in A$. Let $\theta$ be a polarization. Let $\theta_x$ denote its translate by $x$. Let $0\leq k\leq g$. If we have $$ \theta^{g-k}(\theta_x-\theta)^k=0 $$ in $\text{Ch}(A)$, where the exponents are intersection products, then is it true that $(\theta_x-\theta)^k=0$?
For context, a formula of Beauville gives us that $$ \frac{\theta^{g-k}}{(g-k)!}(\theta_x-\theta)^k=\frac{\theta^g}{g!}*\gamma(x)^{*k} $$ where $*$ is the Pontryagin product and $\gamma(x)=\sum_{v=1}^{g}\frac{1}{v}([0]-[x])^{*v}$. I want to show that if $\gamma(x)^{*k}=0$, then $(\theta_x-\theta)^k=0$. This comes up in the following preprint of Voisin in lemma 0.7: https://arxiv.org/pdf/1802.07153.pdf
Your question has an affirmative answer although I could not locate it exactly, either in Bloch's paper or in Beauville's paper.
You can prove it directly by dominating your abelian variety $A$ by a product $C^g$---mapping to $A$ by the sum map. Then the two conditions are equivalent to the fact that $(\theta_x - \theta)^k$ pullback to $0$ in $\text{CH}_0(C^k)$, where $C^k$ maps to $A$ by the sum map.