Prove that there are no real numbers $a,b,c,d$ such that this system of equations holds. Prove that there are no real numbers $a,b,c,d$ such that this system of equations holds.
$$a^3+c^3=2$$
$$a^2b+c^2d=0$$
$$b^3+d^3=1$$
$$ab^2+cd^2=-6$$
I think that we have to use the fact that $x^2-xy+y^2\ge0$ as this is a two part question with the a) part being to prove that $x^2-xy+y^2\ge0$ which is a simple $AG$ equation. Here's some of what I tried.
I think it's important to figure out
$d^3-cd^2+c^2d=d(d^2-cd+c^2)$
$c^3-c^2d+cd^2=c(c^2-cd+d^2)$
$b^3-ab^2+a^2b=b(b^2-ab+a^2)$
$a^3-a^2b+ab^2=a(a^2-ab+b^2)$
but I'm not sure how to effectively use this as summing these four terms together we get $a^3+b^3+c^3+d^3$ which equals $3$ which I don't see a use for. But I'm 99% sure we do have to use them in some way. I've been stuck at this problem for the past hour or so with no significant progression so any help's appreciated! Thanks in advance.
 A: Note that $d=0$ is impossible by second and fourth equation. So we can substitute $c=-(6+ab^2)/d^2$. Taking discriminants we obtain
$$c= \frac{d(431d^3 - 323)}{3870}$$ and then that
$$
431d^6 - 431d^3 + 11664=0
$$
It has no real solution.
Edit: This confirms the claim, but certainly there is some "nicer" way to see this.
A: Here is a simpler argument more in line with the OP's thinking. That is, using the fact that
$$
x^3+y^3=(x+y)(x^2-xy+y^2),
$$
where the quadratic factor on the right is always non-negative.

Key observation: $x^3+y^3$ and $x+y$ have the same sign for all real numbers $x,y$.

Assume contrariwise that there exists a solution $(a,b,c,d)\in\Bbb{R}^4$.

*

*The first equation implies that $a+c>0$.

*The third equation implies that $b+d>0$.

*Then consider the linear combination of all the equations where the 1st and the 3rd equation get weight $1$ and the 2nd the 4th get multiplied by $3$. We arrive at
$$(a^3+3a^2b+3ab^2+b^3)+(c^3+3c^2d+3cd^2+d^3)=2+3\cdot0+3\cdot(-6)+1=-15.$$ In other words
$$(a+b)^3+(c+d)^3=-15.$$

*The previous equation implies that $(a+b)+(c+d)<0$. This is impossible in light of items 1 and 2.


To an extent this exposes that last equation as the cause of non-existence of solutions. If we replace $-6$ with a non-negative constant it is possible to find real solutions. I don't know what the exact threshold is. One of the resident experts on inequalities might be able to solve that.
A: This system is equivalent to a system containing the equation $11664 - 431 d^3 + 431 d^6=0$ which has no real roots (the discriminant of $431y^2-431y+11664$ is negative).
A: The big hammer here is to use Gröbner bases as a way of eliminating the variables one by one, and ending up with a system equivalent to he given one so that at least one of the equations contains only a single variable. There are many possible choices relying on selecting an order of the elements.
Finding a Gröbner basis with pencil and paper alone is possible (Buchberger's algorithm) but tedious. I asked Mathematica for help, and it spat out:
In[1]:= GroebnerBasis[{a^3 + c^3 - 2, a^2 b + c^2 d, b^3 + d^3 - 1, a b^2 + c d^2 + 6}, {a, b, c, d}]
Out[1]:= {11664 - 431 d^3 + 431 d^6, 3870 c + 323 d - 431 d^4, -1 + b^3 + d^3, 3870 a - 108 b + 431 b d^3}

Among other things this implies that for any solution $(a,b,c,d)$ of the given system, the last variable $d$ must satisfy the equation
$$
11664 - 431 d^3 + 431 d^6=0.\qquad(*)
$$
The quadratic formula says that the equation $11664 - 431 x + 431 x^2=0$ has no real zeros. Therefore there are no real zeros of $(*)$ either.

From the Gröbner basis we see that there are 18 complex solutions $(a,b,c,d)$. The first sextic gives six different values for $d$. Thene the second element of the Gröbner basis shows the dependence of $c$ on $d$ (fully determined). Then the third entry on the Gröbner basis tells us that to each $d$ there are $3$ distinct choices of $b$. Finally, the last entry of the Gröbner basis tells us that to given $b$ and $d$ there is a single possibility for $a$. A total of $18$ combinations.

This is, of course, highly unsatisfactory. Anyway, it could be possible to derive equation $(*)$ with a few tricks.
