An algebra problem by Romeo Dane Melendrez

Algebra Level 3

If a 2 + b 2 + c 2 = 2 a + 4 b + 6 c − 14 a^2 + b^2 + c^2 = 2a + 4b + 6c - 14

Find the value of a 3 + b 3 + c 3 a^3 + b^3 + c^3 .


The answer is 36.

Romeo Dane Melendrez by Romeo Dane Melendrez , 21, Philippines

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1 solution

We move everything to one side, we get:

a 2 + b 2 + c 2 − 2 a − 4 b − 6 c + 14 = 0 a^2 + b^2 + c^2 - 2a - 4b - 6c + 14 = 0

( a 2 − 2 a + 1 ) + ( b 2 − 4 b + 4 ) + ( c 2 − 6 c + 9 ) = 0 (a^2 - 2a + 1) + (b^2 - 4b + 4) + (c^2 - 6c + 9) = 0

( a − 1 ) 2 + ( b − 2 ) 2 + ( c − 3 ) 2 = 0 (a - 1)^2 + (b - 2)^2 + (c - 3)^2 = 0

therefore: a = 1, b = 2, and c = 3

then a 3 + b 3 + c 3 = 1 3 + 2 3 + 3 3 = 36 a^3 + b^3 + c^3 = 1^3 + 2^3 + 3^3 = 36

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Are you assuming that a, b, c are real?

Calvin Lin Staff - 3 years, 8 months ago

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