a3+b3+c3-3abc=0,if a+b+cnot equal to 0
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If “ a3+b3+c3=3abc then find the value of (a+b+c) where a≠b≠c”?
Value of (a+b+c) is zero.
Why?
we know,
a3 + b3 + c3 - 3abc = (a + b + c )(a2 + b2 + c2 -ab - ac -bc)
Now it is given that : a3 + b3 + c3 = 3abc
So,
a3 + b3 + c3 - 3abc = 0
(a + b + c )(a2 + b2 + c2 -ab - ac -bc) = 0
this means either
(a2 + b2 + c2 -ab - ac -bc) = 0 or (a + b + c ) = 0
(a2 + b2 + c2 -ab - ac -bc) = 0 cannot be zero because:
2a² + 2b² + 2c² -2ab - 2ac - 2bc = 0
a² + b² -2ab + a² +b² +2c² - 2ac -2bc = 0
(a-b)² + a² + c² -2ac + b² + c² -2bc = 0
(a-b)² + (a-c)² + (b-c)² = 0
(a-b)² , (a-c)² ,(b-c)² >=0
As a≠b≠c , The given value cannot be zero.
This means (a + b + c ) has to be zero
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