Question

if a+2b+3c=0, prove that a³+8b³+27c³=18abc

Answer 1

a+2b+3c=0 then
a+2b=-3c
(a+2b)3=(-3c)3 (By cubing)
a3+(2b)3+3(a)(2b)(a+2b) = -27c3
a3+8b3+6ab (-3c) = -27c3 [a+2b=-3c]
a3+8b3-18abc = -27c3
a3+8b3+27c3 = 18abc

Answer 2

Answer:

If a+2b+3c=0 and a³+2b³+3c³=18abc

Then,

a³+2b³+3c³-18abc;

(a³+b³+c³-3abc)=(a+b+c)(a²+b²+c²-ab-bc-ca)

After this we have to solve the equation by using this identity and prove it!

Hope you like the answer!

Step-by-step explanation: