Giải thích các bước giải:

$\begin{aligned}
&\text{Ta có:} \\
&ac = b^2 \\
&\dfrac{a}{b} = \dfrac{b}{c} \\
&bd = c^2 \\
&\dfrac{b}{c} = \dfrac{c}{d} \\
&\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} \\
&\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = \dfrac{2a}{2b} = \dfrac{3b}{3c} \\
&\dfrac{a}{b} = \dfrac{2a}{2b} = \dfrac{3b}{3c} = \dfrac{c}{d} = \dfrac{2a + 3b - c}{2b + 3c - d} \\
&\left( \dfrac{a}{b} \right)^3 = \left( \dfrac{2a + 3b - c}{2b + 3c - d} \right)^3 \\
&\left( \dfrac{a}{b} \right)^3 = \dfrac{a}{b} \cdot \dfrac{a}{b} \cdot \dfrac{a}{b} \\
&\left( \dfrac{a}{b} \right)^3 = \dfrac{a}{b} \cdot \dfrac{b}{c} \cdot \dfrac{c}{d} \\
&\left( \dfrac{a}{b} \right)^3 = \dfrac{a \cdot b \cdot c}{b \cdot c \cdot d} \\
&\left( \dfrac{a}{b} \right)^3 = \dfrac{a}{d} \\
&\dfrac{a}{d} = \left( \dfrac{2a + 3b - c}{2b + 3c - d} \right)^3 \\
\end{aligned}$