Question

If a + b + c = 0, show that : a3 + b3 + c3 = 3abc.​

Answer 1

Answer:

Given below...

Step-by-step explanation:

We have,

a+b+c= 0

Then, a+b= -c. (1)

Cube on both the sides,

(a+b)³=(-c)³

You know that, (a+b)³= a³+b³+3ab(a+b)

a³+b³+3ab(a+b)= (-c)³

Put (a+b) value as (-c) from the (1)

a³+b³-3ab(-c)= -c³

a³+b³-3abc=-c³

a³+b³+c³= 3abc

I hope it will help you out...

Answer 2

Step-by-step explanation:

a+B+c= 0

a+b+= -c

cubing on both sides

(a+b) ^3 = (-c)^ 3

a^3+ b^3+3ab(a+b)= -c^3

a^3+ b^3+3ab(-c)= -c^3

a^3+ b^3-3abc= -c^3

a^3+ b^3+c^3=3abc