Given: \(a \propto b\) ⇒ \(a = k_1b\), where \(k_1\) is a non-zero constant Also, \(b \propto c\) ⇒ \(b = k_2c\), where \(k_2\) is a non-zero constant So, \(a = k_1b = k_1k_2c\) Now consider: \(\cfrac{a^3 + b^3 + c^3}{abc}\) \(= \cfrac{(k_1k_2c)^3 + (k_2c)^3 + c^3}{k_1k_2c \cdot k_2c \cdot c}\) \(= \cfrac{c^3(k_1^3k_2^3 + k_2^3 + 1)}{k_1k_2^2c^3}\) \(= \cfrac{k_1^3k_2^3 + k_2^3 + 1}{k_1k_2^2}\) Which is a non-zero constant. ∴ \(a^3 + b^3 + c^3 \propto abc\(Proved)