\(a \propto b\) ⇒ \(a = k_1b\) [where \(k_1\) is a non-zero constant] Again, \(b \propto c\) ⇒ \(b = k_2c\) [where \(k_2\) is a non-zero constant] Therefore, \(a = k_1b = k_1k_2c\) Now, \(\cfrac{a^3 + b^3 + c^3}{5abc}\) \(= \cfrac{(k_1k_2c)^3 + (k_2c)^3 + c^3}{5 \cdot k_1k_2c \cdot k_2c \cdot c}\) \(= \cfrac{c^3(k_1^3k_2^3 + k_2^3 + 1)}{5k_1k_2^2c^3}\) \(= \cfrac{k_1^3k_2^3 + k_2^3 + 1}{5k_1k_2^2}\) \(=\) a non-zero constant \(\therefore a^3 + b^3 + c^3 \propto 5abc\) (Proved)