<s>[INST] Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$

Show this equation holds by squaring both sides and comparing terms up to $x^3$.

I wonder, how can I square the right hand side? 
 [/INST] Hint:
\begin{align}
(a + b + c + \ldots )^2  & =  (a+b+c+\ldots) \times (a+b+c+\ldots)   \\
& = a^2 + ba + ca + \ldots + a b + b^2 + cb + \ldots ac + bc +c ^2 + \ldots 
\end{align}
Can you take it from here?
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