Under what condition ae +b3+ c3= 9abc
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Answer:
27a³-b³+c³+9abc
=>(3a)³+(-b)³+(c)³-3(3a)(-b)(c)
=>(3a-b+c)(9a²+b²+c²+3ab+bc-3ca)
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Answer. The possible values are the nonnegative integers that are either divisible by 9 or not divisible by 3. ... To show that no other values are possible, first note that by the AM-GM inequality, for nonnegative A,B,C we have 1 3 (A3 + B3 + C3) ≥ 3 √ A3B3C3 = ABC, and therefore f(A,B,C) ≥ 0.
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