In general: \((a-b)^3=a^3-3a^2b+3ab^2-b^3\) so we can also do it like this:
\((2x-\frac{3}{4}y^2)^3\rightarrow (2x)^3-3(2x)^2(\frac{3}{4}y^2)+3(2x)(\frac{3}{4}y^2)^2-(\frac{3}{4}y^2)^3\\ \rightarrow 8x^3-9x^2y^2+\frac{27}{8}xy^4-\frac{27}{64}y^6\)
where \(a=2x \text{ and } b=\frac{3}{4}y^2\)
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