Prove that a = 2 , b = 2 , c = 2 for a system of three equations

rigliztetbf

rigliztetbf

Answered question

2022-06-12

Prove that a = 2 , b = 2 , c = 2 for a system of three equations.
Let us consider a system equations :
(i) a 3 + b = 3 a + 4 (ii) 2 b 3 + c = 6 b + 6 (iii) 3 c 3 + a = 9 c + 8
I have tried with the following steps :
6 × ( i ) + 3 × ( i i ) + 2 × (iii) gives:
6 ( a 3 + b 3 + c 3 ) + 6 b + 3 c + 2 a = 18 ( a + b + c ) + 58.
But I don't know how to solve in this way. By inspection method I have seen a = b = c = 2.

Answer & Explanation

Korotnokby

Korotnokby

Beginner2022-06-13Added 19 answers

Write the equations as
( a 2 ) ( a + 1 ) 2 = 2 b 2 ( b 2 ) ( b + 1 ) 2 = 2 c 3 ( c 2 ) ( c + 1 ) 2 = 2 a
Note that ( a , b , c ) = ( 2 , 2 , 2 ) is a valid solution.
a > 2 b < 2 c > 2 a < 2.. A similar logic precludes a < 2. This forces a = b = c = 2 which is the unique solution
Emmy Dillon

Emmy Dillon

Beginner2022-06-14Added 5 answers

This is ugly, but you can rewrite the system as.
b = a 3 + 3 a + 4 c = 2 b 3 + 6 b + 6 a = 3 c 3 + 9 c + 8
Now, you can substitute the first into the second and that result into the third.
This will show one of the roots, a = 2 immediately and another useless one.
You can then sub back in the other two and derive the other two roots.

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