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 Therefore, \(a = k_1b = k_1k_2c\) Now consider: \(\frac{a^3 + b^3 + c^3}{3abc}\) \(= \frac{(k_1k_2c)^3 + (k_2c)^3 + c^3}{3 \cdot k_1k_2c \cdot k_2c \cdot c}\) \(= \frac{c^3(k_1^3k_2^3 + k_2^3 + 1)}{3k_1k_2^2c^3}\) \(= \frac{k_1^3k_2^3 + k_2^3 + 1}{3k_1k_2^2}\) This is a non-zero constant. ∴ \(a^3 + b^3 + c^3 \propto 3abc\) (Proved)