Ta có:\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow1+\dfrac{b}{a}=1+\dfrac{d}{c}\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
\(\Rightarrow\left(\dfrac{a+b}{a}\right)^3=\left(\dfrac{c+d}{c}\right)^3\Rightarrow\dfrac{a^3}{\left(a+b\right)^3}=\dfrac{c^3}{\left(c+d\right)^3}\)