Question

when a=2, b=0, c=-2. Find the value of a3+b3+c3-3abc
please give the answer as soon as possible​

Answer 1

Answer:

We know from the identity,

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

i.e.

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

[∴ a + b + c = 0]

a3+b3 + c3 – 3abc = 0

So

a³+b³+c³=3abc

Answer 2

Answer:

0

Step-by-step explanation:

Given :-

a=2, b=0, c=-2

Find :-

a3+b3+c3-3abc

Solution :-

=>2×3+0×3+-2×3-3×2×0×-2

=>6+0-6-0

=>0

Therefore, the answer is 0.