A system of huge equation

Algebra Level 5

Find all the possible sum of a , b , c ∈ R a,b,c \in \mathbb{R} satisfying the following system of equation

{ a 2 = b 3 − 18 3 b = c 3 − 52 3 c b 2 = a 3 + 26 3 a = c 3 + 44 3 c c 2 = a 3 + 47 3 a = b 3 + 117 3 b \begin{cases} { a }^{ 2 }=\dfrac { { b }^{ 3 }-18 }{ 3b } =\dfrac { { c }^{ 3 }-52 }{ 3c } \\ { b }^{ 2 }=\dfrac { { a }^{ 3 }+26 }{ 3a } =\dfrac { { c }^{ 3 }+44 }{ 3c } \\ { c }^{ 2 }=\dfrac { { a }^{ 3 }+47 }{ 3a } =\dfrac { { b }^{ 3 }+117 }{ 3b } \end{cases}

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The answer is 8.

Isaac Yiu Math Studio by Isaac YIU Math Studio , 15, Hong Kong

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2 solutions

David Vreken
Aug 2, 2019

a 2 = b 3 − 18 3 b a^2 = \frac{b^3 - 18}{3b} and b 2 = a 3 + 26 3 a b^2 = \frac{a^3 + 26}{3a} rearrange to b 3 − 3 a 2 b = 18 b^3 - 3a^2b = 18 and a 3 − 3 a b 2 = − 26 a^3 - 3ab^2 = -26 , and using the identity ( b 3 − 3 a 2 b ) 2 + ( a 3 − 3 a b 2 ) 2 = ( a 2 + b 2 ) 3 (b^3 - 3a^2b)^2 + (a^3 - 3ab^2)^2 = (a^2 + b^2)^3 this gives us a 2 + b 2 = 1 8 2 + ( − 26 ) 2 3 = 10 a^2 + b^2 = \sqrt[3]{18^2 + (-26)^2} = 10 or

a 2 + b 2 = 10 a^2 + b^2 = 10

By a similar argument,

a 2 + c 2 = 17 a^2 + c^2 = 17

b 2 + c 2 = 25 b^2 + c^2 = 25

These three equations solve to a 2 = 1 a^2 = 1 , b 2 = 9 b^2 = 9 , and c 2 = 16 c^2 = 16 , and from the original equations we can deduce that a = 1 a = 1 , b = 3 b = 3 , and c = 4 c = 4 , so that a + b + c = 8 a + b + c = \boxed{8} .

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That's my solution XD I have tried to type this but I suddenly delete all things ;(

Isaac YIU Math Studio - 1 year, 10 months ago

There are three solutions to this set of equations. Only one set is real numbers; the other two are complex.

{{1,3,4},{-(-1)^(1/3),-3 (-1)^(1/3),-4 (-1)^(1/3)},{(-1)^(2/3),3 (-1)^(2/3),4 (-1)^(2/3)}}

0 Helpful 0 Interesting 0 Brilliant 1 Confused

Can u explain it?

Isaac YIU Math Studio - 1 year, 10 months ago

This is an over-determined, but consistent, set of equations. For example, the a 2 a^2 set is unnecessary.

Using the other two sets and eliminating the a a variable from them gives 4 b = 3 c ∧ c 3 = 64 4 b=3 c\land c^3=64 .

Since we are only looking for real solutions, that immediately gives b=3 and c=4.

Using b 2 = a 3 + 26 3 a b^2=\frac{a^3+26}{3 a} and substituting b = 3 b=3 gives 9 = a 3 + 26 3 a 9=\frac{a^3+26}{3 a} or a 3 = 1 a^3=1 or a = 1 a=1 . Q. E. D.

A Former Brilliant Member - 1 year, 10 months ago

How can you get 4 b = 3 c 4b=3c and c 3 = 64 c^3=64

Isaac YIU Math Studio - 1 year, 10 months ago

Basically, see Mr. Vreken's solution. The methodology is similar. One difference is that I retained the cubes into that step to show that the answer is actually complex.

A Former Brilliant Member - 1 year, 10 months ago

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