3a+5b+4c=0 Let 3a=A, 5b=B, and 4c=C.
The given equation can be restated: A+B+C=0 ··· Eq.1 Use a standard polynomial form:
A³+B³+C³-3ABC=(A+B+C)(A²+B²+C²-AB-BC-CA) ··· Eq.2 Substitute Eq.1 in Eq.2. Eq.2 can be restated:
A³+B³+C³=3ABC ··· Eq.3 Here, A³=(3a)³=27a³, B³=(5b)³=125b³, and C³=(4c)³=64c³, so we have:
A³+B³+C³=27a³+125b³+64c³ ··· Eq.4
While, 3ABC=3(3a)(5b)(4c)=180abc ··· Eq.5 From Eq.3, Eq.4, and Eq.5, we have:
27a³+125b²+64c³=180abc Q.E.D.