Super hard system of equations 
Solve the system of equation for real numbers

\begin{split}
(a+b)    &(c+d)     &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}

First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(bc+ad)^2$$
Use this identity to (4) too
and simplify (3),
we obtain $$(a^2+b^2-ab)(c^2+d^2-cd)=7$$
And suppose $x=abcd$ use $ac=x /bd , bc=x/ad$
But got stuck...

Not really complete, but an interesting result using simple algebraic manipulations.

Write:
$$\begin{align}
a^2+b^2&=(a+b)^2-2ab\\
a^3+b^3&=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)\\
a^4+b^4&=((a^2)^2+(b^2)^2)=\cdots=(a+b)^4+2a^2b^2-4ab(a+b)^2\\
\vdots
\end{align}$$
From $(2)$, we have:
$$\begin{align}
(a^2+b^2)(c^2+d^2)&=((a+b)^2-2ab)((c+d)^2-2cd)=9\\
&=\color{red}{(a+b)^2(c+d)^2}-2ab(c+d)^2-2cd(a+b)^2-4abcd=9
\end{align}$$
But from $(1)$, we know that $(a+b)(c+d)=1$, then the text in $\color{red}{\text{red}}$ is also equal to $1$, so the result above becomes:
$$2abcd-ab(c+d)^2-cd(a+b)^2=4\tag{1*}$$
From $(3)$, we have:
$$\begin{align}
(a^3+b^3)(c^3+d^3)&=\color{red}{(a+b)}((a+b)^2-3ab)\color{red}{(c+d)}((c+d)^2-3cd)=7\\
&=((a+b)^2-3ab)((c+d)^2-3cd)=7\\
&\qquad\vdots\\
&=3abcd-ab(c+d)^2-cd(a+b)^2=2\tag{2*}
\end{align}$$
Adding $(1*)$ and $(2*)$, we get $abcd=-2$ and that $\color{pink}{ab(c+d)^2+cd(a+b)^2=-8}$.

Now $(4)$ is really tricky, but you can write it as:
$$\begin{align}
\left((a+b)^4+2a^2b^2-4ab(a+b)^2\right)\left((c+d)^4+2c^2d^2-4cd(c+d)^2\right)&=25
\end{align}$$
Expanding, and we can eliminate $(a+b)^4(c+d)^4$ since it is equal to $1$. Then we have:
$$\begin{align}
a^2b^2(c+d)^4-2ab(c+d)^2+c^2d^2(a+b)^4+2(abcd)^2-4abc^2d^2(a+b)^2-2cd(a+b)^2-4a^2b^2cd(c+d)^2-4abcd=12
\end{align}$$
Using the fact that $abcd=-2$, then we can shorten the equation above into:
$$\color{red}{a^2b^2(c+d)^4+c^2d^2(a+b)^4}+5ab(c+d)^2-10cd(a+b)^2+16=12\tag{3*}$$
However, you can see that the text in $\color{red}{\text{red}}$ looks very close to the square of two sums:
$$\color{red}{a^2b^2(c+d)^4+c^2d^2(a+b)^4}=(ab(c+d)^2+cd(a+b)^2)^2-2\color{blue}{abcd(a+b)^2(c+d)^2}$$
However we already know the value of the part in $\color{blue}{\text{blue}}$ to be $-2
\cdot 1$.
Now we can write $(3*)$ as:
$$(ab(c+d)^2+cd(a+b)^2)^2+6ab(c+d)^2-10cd(a+b)^2=-8$$

From here, you can substitute $x=ab(c+d)^2$ and $y=cd(a+b)^2$, which gives two systems of equation:
$$(x+y)^2+6x-10y=-8\\
x+y=-8$$
This has one solution:
$$x=-\frac{19}2\,\,y=\frac32$$

You can try working from here.
