An algebra problem by Aziz Alasha

Algebra Level 3

Given that a + b + c = 0 a + b + c = 0 , where a a , b b , and c c are non-zero real numbers, then:

a 3 + b 3 + c 3 a b c = ? \large \frac{{a}^3 + {b}^3 + {c}^3}{abc} = ?


The answer is 3.

Aziz Alasha by Aziz Alasha , 59, Sudan

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1 solution

Note first that in general

( a + b + c ) 3 = a 3 + b 3 + c 3 + 3 ( a 2 b + a 2 c + b 2 a + b 2 c + c 2 a + c 2 b ) + 6 a b c = a 3 + b 3 + c 3 + 3 ( ( a + b + c ) ( a b + b c + c a ) − 3 a b c ) + 6 a b c (a + b + c)^{3} = a^{3} + b^{3} + c^{3} + 3(a^{2}b + a^{2}c + b^{2}a + b^{2}c + c^{2}a + c^{2}b) + 6abc = a^{3} + b^{3} + c^{3} + 3((a + b + c)(ab + bc + ca) - 3abc) + 6abc .

With a + b + c = 0 a + b + c = 0 this simplifies to 0 = a 3 + b 3 + c 3 + 3 ( 0 − 3 a b c ) + 6 a b c = a 3 + b 3 + c 3 − 3 a b c ⟹ a 3 + b 3 + c 3 a b c = 3 0 = a^{3} + b^{3} + c^{3} + 3(0 - 3abc) + 6abc = a^{3} + b^{3} + c^{3} - 3abc \Longrightarrow \dfrac{a^{3} + b^{3} + c^{3}}{abc} = \boxed{3} .

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