In my notes, it says if $R$ is a ring and $a\in R$ then ${\langle a \rangle}^3 \subseteq RaR$.
I cannot see where this comes from as $RaR=\{r_1as_1 + \ldots + r_nas_n: n\in\mathbb{N}, r_i\in R, s_i\in R\}$ whereas ${\langle a \rangle}^3=\{x_1y_1z_1 + \ldots + x_ny_nz_n: n\in\mathbb{N}, x_i\in {\langle a \rangle}, y_i\in {\langle a \rangle},z_i \in {\langle a \rangle}\}$ if I am not mistaken.
Would someone be able explain to explain why this is a subset of $RaR$ please ?
Because $R$ is not necessarily commutative and not necessarily unital, the general element of $\langle a\rangle$ has the form $$na + ra + as + \sum_{i=1}^m r_ias_i$$ where $n\in\mathbb{Z}$, $m$ is a nonnegative integer, and $r,s,r_i,s_i\in R$.
Let us take a product of three such elements, $$\begin{align*} x &= n_xa + r_xa + as_x + \sum_{i=1}^{m_x}r_{i,x}as_{i,x}\\ y &= n_ya + r_ya + as_y + \sum_{i=1}^{m_y}r_{i,y}as_{i,y}\\ z &= n_za + r_za + as_z + \sum_{i=1}^{m_z}r_{i,z}as_{i,z} \end{align*}$$ Now, imagine you multiply them out. It’s going to be a bit of a mess, as it will have 64 terms and I don’t want to write them all out. But for example, let’s say we look at the term given by the product of $r_xa$, $\sum_{i=1}^{m_y}r_{i,y}as_{i,y}$, and $n_za$. This term will be $$\begin{align*} r_xa\left(\sum_{i=1}^{m_y}r_{i,y}as_{i,y}\right)n_za &= n_z\left((r_xa)\left(\sum_{i=1}^{m_y}r_{i,y}as_{i,y}\right) a\right)\\ &= n_z\left( \sum_{i=1}^{m_y}r_xar_{iy}as_{i,y}a\right). \end{align*}$$ But $r_xar_{iy}as_{i,y}a = (r_xar_{iy})a(s_{i,y}a)\in RaR$, and hence the sum lies in $RaR$, and hence the sum of $n_z$ copies of that element lies in $RaR$.
Convince yourself that each of those 64 products is going to be a sum of elements of the form $vaw$ for some $v,w\in R$.