\(a ∝ b, b ∝ c\) হলে দেখাই যে, \(a^3 b^3+b^3 c^3+c^3 a^3\) \(∝abc(a^3+b^3+c^3 )\)
\(\because a\propto b, \therefore a=k_1b=k_1k_2c\)
\(\because b\propto c, \therefore b=k_2c\) [\(k_1, k_2\) অশূন্য ভেদ ধ্রুবক ]
\(\therefore \cfrac{a^3b^3+b^3c^3+c^3a^3}{abc(a^3+b^3+c^3)}\)
\(=\cfrac{k_1^3k_2^3c^3.k_2^3c^3+k_2^3c^3.c^3+c^3.k_1^3k_2^3c^3}{k_1k_2c.k_2c.c(k_1^3k_2^3c^3+k_2^3c^3+c^3)}\)
\(=\cfrac{c^6(k_1^3k_2^6+k_2^3+k_1^3k_2^3}{c^3.k_1k_2^2.c^3(k_1^3k_2^3+k_2^3+1)}\)
\(=\cfrac{(k_1^3k_2^6+k_2^3+k_1^3k_2^3}{k_1k_2^2(k_1^3k_2^3+k_2^3+1)}\)
\(=\) ধ্রুবক
\(\therefore a^3b^3+b^3c^3+c^3a^3\propto abc(a^3+b^3+c^3)\) (প্রমাণিত)