Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Từ đó, ta được:\(\dfrac{\left(a+c\right)^3}{\left(b+d\right)^3}=\dfrac{\left(bk+dk\right)^3}{\left(b+d\right)^3}=\dfrac{\left[k\left(b+d\right)\right]^3}{\left(b+d\right)^3}=\dfrac{k^3.\left(b+d\right)^3}{\left(b+d\right)^3}=k^3\left(1\right)\) \(\dfrac{\left(a-c\right)^3}{\left(b-d\right)^3}=\dfrac{\left(bk-dk\right)^3}{\left(b-d\right)^3}=\dfrac{\left[k\left(b-d\right)\right]^3}{\left(b-d\right)^3}=\dfrac{k^3.\left(b-d\right)^3}{\left(b-d\right)^3}=k^3\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{\left(a+c\right)^3}{\left(b+d\right)^3}=\dfrac{\left(a-c\right)^3}{\left(b-d\right)^3}\)
