If `a^(1/3)+\ b^(1/3)+\ c^(1/3)=0` then
(a)`a+b+c=0` (b) `(a+b+c)^3=27\ a b c` (c)`a+b+c=3a b c` (d) `a^3+b^3+c^3=0`

If a^((1)/(3))+b^((1)/(3))+c^((1)/(3))=0 then (a+b+c)^(3)=

If a^(1//3)+b^(1//3)+c^(1//3)=0 , then show that (a+b+c)^(3)=27 abc .

If a ^(1//3) + b ^(1//3) + c ^(1//3) = 0, then the value of (a + b + c) ^(3) will be:

If a + b + c = 0 , show that a^(3) + b^(3) + c^(3) = 3abc The following are the steps involved in showing the above result. Arrange them in sequential order (A) a^(3) + b^(3) + 3ab (-c) = -c^(3) (B) (a + b)^(3) = (-c)^(3) (C) a + b + c = 0 rArr a + b = -c (D) a^(3) + b^(3) + 3ab (a +b) = -c^(3) (E) a^(3) + b^(3) + c^(2) = 3abc

Prove that |(a+b, b, c), (b+c, c, a), (c+a, a, b)| = 3abc - a^3-b^3-c^3.

If a+b+c=0 then (a^3+b^3+c^3) is

If a^(3) + b^(3) + c^(3)=3abc and a, b , c are distinct numbers. Which option is correct? (a) a+b+c= 0 (b) a= b= c