I want to show that
$$\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)=\operatorname{Ass}_R\left(\dfrac{R}{x^3R}\right)$$
where $x\in R$ is a non-zero-divisor and non-unit element and $R$ is a Noetherian ring. To proof this exercise, I've been used these two short exact sequences:
$$0\to\dfrac{R}{x^2R}\to\dfrac{R}{x^3R}\to\dfrac{R}{xR}\to0$$
and
$$0\to\dfrac{R}{xR}\to\dfrac{R}{x^3R}\to\dfrac{R}{x^2R}\to0.$$
From the first sequence we have:
$$\operatorname{Ass}_R\left(\dfrac{R}{x^2R}\right)\subseteq\operatorname{Ass}_R\left(\dfrac{R}{x^3R}\right)\subseteq\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)\cup\operatorname{Ass}_R\left(\dfrac{R}{x^2R}\right)=\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)$$
and second one gives us:
$$\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)\subseteq\operatorname{Ass}_R\left(\dfrac{R}{x^3R}\right)\subseteq\operatorname{Ass}_R\left(\dfrac{R}{xR}\right)\cup\operatorname{Ass}_R\left(\dfrac{R}{x^2R}\right)=\operatorname{Ass}_R\left(\dfrac{R}{x^2R}\right).$$ So we have the equality.
If you see Page 289 of the book, by the third isomorphism theorem, $xR/x^nR\cong R/x^{n-1}R$, $0→R/x^{n-1}R→R/x^nR→R/xR→0$.
From the sequence, you can get the result inductively. Let $n=2$. Then $0→R/xR→R/x^2R→R/xR→0$ and $Ass_{R}(R/xR)\subset Ass_{R}(R/x^2R)\subset Ass_{R}(R/xR)∪ Ass_{R}(R/xR)=Ass_{R}(R/xR)$. And let $n=3$....