Important Identity
Algebra
Level
2
if = 3abc and a,b,c are positive numbers, then
a>b>c
a=b=c
a<b<c
none of above
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1 solution
a 3 + b 3 + c 3 = 3abc
a 3 + b 3 + c 3 - 3abc = 0.
(a + b + c )( a 2 + b 2 + c 2 -ab - ac -bc) = 0
here (a + b + c) cannot be equal to zero since a,b,c are positive. Therefore ( a² +b² +c²-ab - ac -bc) = 0.
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
since postive numbers cannot be added to make zero, each bracket should be zero a-b = 0 a-c = 0 b-c =0 a = b a= c b = c a = b = c