MHB Find all real solutions of a, b and c

[anemone]

Find all real solutions $a,\,b,\,c$ of the system of equations below:

$a^3+b=3a+4$

$2b^3+c=6b+6$

$3c^3+a=9c+8$

 

[Opalg]

Find all real solutions $a,\,b,\,c$ of the system of equations below:

$a^3+b=3a+4$

$2b^3+c=6b+6$

$3c^3+a=9c+8$

[sp]The "obvious" solution is $a=b=c=2$. In fact, if $a=2$ then the first equation says that $b=2$. The second equation then says that $c=2$, and the third equation confirms that $a=2$.

Now suppose that there is another solution with $a>2$. Then the first equation says that $b<2$. The second equation then says that $c>2$, and the third equation says that $a<2$. But that contradicts the assumption that $a>2$, so there are no such solutions.

Similarly, if there is a solution with $a<2$ then the first equation says that $b>2$. The second equation then says that $c<2$, and the third equation says that $a>2$. Again, that contradicts the assumption that $a<2$, so there are no such solutions.

Therefore there is only the one solution.[/sp]

 

[anemone]

This is definitely an eye-opening solution for me...thank you so much Opalg for your great solution and thanks for participating!

Algebraic solution of other:

Spoiler

From $a^3+b=3a+4$, rewrite it so we have $a^3-1-1-3a=2-b\,\implies\,(a-2)(a-1)^2=2-b$---(1)

From $2b^3-2-2-6b=2-c$, rewrite it such that $2(b-2)(b+1)^2=(2-c)$---(2) and

$3c^3-3-3-9c=2-a$ gives $3(c-2)(c+1)^2=(2-a)$---(3)

Multiplying all three equations (1), (2) and (3) yields

$(a-2)(b-2)(c-2)(6(a-1)^2(b+1)^2(c+1)^2+1)=0$

As the last factor is always positive for all real $a,\,b,\,c$, we must have $(a-2)(b-2)(c-2)=0$.

In conjunction with (1), (2), (3), this gives the unique solution $a=b=c=2$.