\(b^2=ac\)
=>\(\dfrac{b}{a}=\dfrac{c}{b}\)
\(c^2=bd\)
=>\(\dfrac{c}{b}=\dfrac{d}{c}\)
=>\(\dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{c}\)
=>\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^3\end{matrix}\right.\)
\(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\dfrac{\left(dk^3\right)^3+\left(dk^2\right)^3-\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3-d^3}\)
\(=\dfrac{d^3k^9+d^3k^6-d^3k^3}{d^3k^6+d^3k^3-d^3}\)
\(=\dfrac{d^3k^3\left(k^6+k^3-1\right)}{d^3\left(k^6+k^3-1\right)}=k^3\)
\(\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
\(=\left(\dfrac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)
\(=\left(\dfrac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right)^3=k^3\)
Do đó: \(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
