+ Ta có
\(k=\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}=k\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=k^3\) (1)
+ Ta có
\(k=\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=k^3\)
=> \(k^3=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\) (2)
=> \(\left(\frac{a+b+c}{b+c+d}\right)^3.\left(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\right)=k^3.k^3=k^6\)