357*. Solve the systems of equations:
a) $\left\{\begin{array}{l}x^{2}+x y+y^{2}=7, \\ y^{2}+y z+z^{2}=13, \\ z^{2}+z x+x^{2}=19 ;\end{array}\right.$
b) $\left\{\begin{array}{l}x y+x z=x^{2}+2, \\ x y+y z=y^{2}+3, \\ x z+y z=z^{2}+4 .\end{array}\right.$
$\triangle$ a) Multiply the first equation of the system by $x-y$, the second by $y-z$, and the third by $z-x$. Why? To obtain a linear equation with $x, y$, and $z$. We will have:

$$
x^{3}-y^{3}=7(x-y), \quad y^{3}-z^{3}=13(y-z), \quad z^{3}-x^{3}=19(z-x)
$$

Add all these equations:

$$
\begin{aligned}
0 & =7 x-7 y+13 y-13 z+19 z-19 x \\
& -12 x+6 y+6 z=0, \quad z=2 x-y
\end{aligned}
$$

Substitute the obtained expression for $z$ into the third equation of the original system:

$$
(2 x-y)^{2}+(2 x-y) x+x^{2}=19, \quad 7 x^{2}-5 x y+y^{2}=19
$$

Now we need to deal with the system of two unknowns $x$ and $y$:

$$
7 x^{2}-5 x y+y^{2}=19, \quad x^{2}+x y+y^{2}=7
$$

Such systems were encountered in section 8.4. Eliminate the free terms by multiplying the first equation by 7 and the second by 19 and subtracting the new equations:

$$
30 x^{2}-54 x y-12 y^{2}=0, \quad 5 x^{2}-9 x y-2 y^{2}=0
$$

We obtained a homogeneous equation. From the original system of equations, it is clear that $y \neq 0$ (think about why). Divide the last equation term by term by $y^{2}$ and let $\frac{x}{y}=t$. Then $5 t^{2}-9 t-2=0$.

Complete the solution on your own.

b) Express the products $x y, y z$, and $z x$ in terms of $x^{2}, y^{2}$, and $z^{2}$ from the system. For this purpose, add all three equations:

$$
\begin{aligned}
& 2 x y+2 x z+2 y z=x^{2}+y^{2}+z^{2}+9 \\
& x y+x z+y z=\frac{1}{2}\left(x^{2}+y^{2}+z^{2}+9\right)
\end{aligned}
$$

Now subtract each of the original system's equations from the last equation:

$$
y z=\frac{1}{2}\left(y^{2}+z^{2}-x^{2}+5\right)
$$

$$
\begin{aligned}
& x z=\frac{1}{2}\left(x^{2}+z^{2}-y^{2}+3\right) \\
& x y=\frac{1}{2}\left(x^{2}+y^{2}-z^{2}+1\right)
\end{aligned}
$$

But what to do with this new system? In each of its equations, eliminate the denominators, collect all terms with unknowns in the left parts, and factorize the left parts:

$2 y z-y^{2}-z^{2}+x^{2}=5$,
$2 x z-x^{2}-z^{2}+y^{2}=3$,
$2 x y-x^{2}-y^{2}+z^{2}=1 ;$
$x^{2}-(y-z)^{2}=5$,
$y^{2}-(x-z)^{2}=3$,
$z^{2}-(x-y)^{2}=1 ;$
$(x+y+z)(x-y+z)=5$,
$(y+x-z)(y-x+z)=3$,
$(z+x-y)(z-x+y)=1$.

Multiply all three equations of the last system:

$$
\begin{aligned}
& (x+y-z)^{2}(x+z-y)^{2}(y+z-x)^{2}=15 \\
& (x+y-z)(x+z-y)(y+z-x)= \pm \sqrt{15}
\end{aligned}
$$

Further, we need to consider two cases depending on the sign in front of $\sqrt{15}$. The plan for further solution is clear.

Answer: a) $(2 ; 1 ; 3),(-2 ;-1 ;-3)$; b) $(2 \sqrt{15} / 3 ; 3 \sqrt{15} / 5 ; 4 \sqrt{15} / 15),(-2 \sqrt{15} / 3$; $-3 \sqrt{15} / 5 ;-4 \sqrt{15} / 15)$

$358 *$. Solve the systems of equations:
a) $\left\{\begin{array}{l}x+y^{2}=z^{3}, \\ x^{2}+y^{3}=z^{4}, \\ x^{3}+y^{4}=z^{5}\end{array}\right.$,
b) $\left\{\begin{array}{l}x(1+y)=z^{2}(1+x), \\ y(1+z)=x^{2}(1+y), \\ z(1+x)=y^{2}(1+z)\end{array}\right.$
c) $\left\{\begin{array}{l}\left(x^{2}+y^{2}\right)(x+y)=15 x y \\ \left(x^{4}+y^{4}\right)\left(x^{2}+y^{2}\right)=85 x^{2} y^{2}\end{array}\right.$
d) $\left\{\begin{array}{l}(x+y)(x+z)=x, \\ (y+z)(y+x)=2 y, \\ (z+x)(z+y)=3 z ;\end{array}\right.$
e) $\left\{\begin{array}{l}x+y+z=0, \\ x^{2}+y^{2}-z^{2}=20, \\ x^{4}+y^{4}-z^{4}=560 ;\end{array}\right.$
f) $\left\{\begin{array}{l}x+y z t=2, \\ y+x z t=2, \\ z+x y t=2, \\ t+x y z=2 .\end{array}\right.$