Prove that $c^3 + a^3 − 2abc = 6b − 11$ Let $a, b, c$ be real numbers such that $(a + b + c)^2 = 3(ab + bc + ca + 1).$ Given,
$$a^3 + b^3 − 2abc = 6c + 2,$$
$$b^3 + c^3 − 2abc = 6a + 9,$$
Prove that $$c^3 + a^3 − 2abc = 6b − 11$$
Note that, subtracting the two equations, we get $$a^3-c^3=6(c-a)+7. $$
Moreover, we have $$a^2+b^2+c^2-ab-bc-ca=3\implies (a-b)^2+(b-c)^2+(c-a)^2=6$$
So, we get $$a^3-c^3=[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7$$ $$\implies a^3-c^3+2c^3-2abc-11=[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7+2c^3-2abc-11.$$
And $$3 a^3 - 2 a^2 b - 4 a^2 c + 2 a b^2 + 4 a c^2 - 2 b^2 c + 2 b c^2 - 3 c^3 - 7 = 0 $$
So, it's enough to show that $$[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7+2c^3-2abc=6b $$
or show that $$2 a^3 - 4 a^2 c + 4 a c^2 + 2 b^3 - 4 b^2 c + 4 b c^2 - 4 c^3=7$$
or show that $$2 (a^3+b^3) - 4 a^2 c + 4 a c^2 - 4 b^2 c + 4 b c^2 - 4 c^3=7$$
I got something ahead, but failed. Any solutions?

Let $s=a+b+c$ and $t=ab+bc+ca$, so that $s^2=3t+3$ is the first given equation. Note that
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=s(s^2-3t)=3s.$$
So,
$$(a^3+b^3-2abc)+(b^3+c^3-2abc)+(c^3+a^3-2abc)=6s.$$
Now simply subtract the other two given equations.
