119. Solve the system of equations:
a) $\left\{\begin{array}{l}y^{4}+x y^{2}-2 x^{2}=0 \\ x+y=6\end{array}\right.$
b) $\left\{\begin{array}{l}x^{2}+y^{2}+x+y=42 \\ x y=15\end{array}\right.$

c) $\left\{\begin{array}{l}y z=a x, \\ z x=b y, \\ x y=c z ;\end{array}\right.$

d) $\left\{\begin{array}{l}\sqrt{(x+3)^{2}}=x+3 \\ \sqrt{(x-3)^{2}}=x^{3}-x\end{array}\right.$
Solution. a) We have:

$$
\begin{gathered}
y^{4}+x y^{2}-2 x^{2}=y^{4}+x y^{2}-x^{2}-x^{2}= \\
=\left(y^{2}-x\right)\left(y^{2}+x\right)+x\left(y^{2}-x\right)= \\
=\left(y^{2}-x\right)\left(y^{2}+2 x\right) .
\end{gathered}
$$

Thus, the first equation of the system decomposes into two equations: $y^{2}-x=0$ and $y^{2}+2 x=0$.

We will solve the system of equations by solving the following two systems:

$$
\left\{\begin{array} { l } 
{ y ^ { 2 } - x = 0 , } \\
{ x + y = 6 }
\end{array} \quad \left\{\begin{array}{l}
y^{2}+2 x=0 \\
y+x=6
\end{array}\right.\right.
$$

Answer: $(4 ; 2),(9 ;-3)$.

b) Multiplying both sides of the second equation by 2 and adding it to the first, we get the system of equations:

$$
\left\{\begin{array}{l}
(x+y)^{2}+(x+y)=72 \\
x y=15
\end{array}\right.
$$

If $x+y=t$, then $t^{2}+t-72=0$, from which $t_{1}=-9, t_{2}=8$.

To find $x$ and $y$, we have two systems of equations:

$$
\left\{\begin{array} { l } 
{ x + y = - 9 , } \\
{ x y = 1 5 ; }
\end{array} \quad \left\{\begin{array}{l}
x+y=8 \\
x y=15
\end{array}\right.\right.
$$

Answer: $(3 ; 5),(5 ; 3),\left(\frac{-9+\sqrt{21}}{2} ; \frac{-9-\sqrt{21}}{2}\right)$,

$$
\left(\frac{-9-\sqrt{21}}{2} ; \frac{-9+\sqrt{21}}{2}\right) \text {. }
$$

c) It is obvious that $x=y=z=0$ is a solution to the given system of equations.

If $x \neq 0, y \neq 0, z \neq 0$, then, multiplying all three equations, we get $x^{2} y^{2} z^{2}=a b c x y z$, i.e., $x y z=a b c$. Substituting the values of $y z, z x$ and $x y$ from the given equations into the left side of this equation, we will have:

$$
\left\{\begin{array}{l}
a x^{2}=a b c \\
b y^{2}=a b c \\
c z^{2}=a b c
\end{array}\right.
$$

i.e., $x^{2}=b c, y^{2}=a c, z^{2}=a b$.

Since the given equations imply that $a, b, c$ are numbers of the same sign, then $b c, a c, a b$ are positive numbers and $x=\pm \sqrt{b c}, y=\pm \sqrt{a c}, z=\pm \sqrt{a b}$.

Answer: $(0 ; 0 ; 0),(\sqrt{b c} ; \pm \sqrt{a c} ; \pm \sqrt{a b}),(-\sqrt{b c} ; \pm \sqrt{a c} ; \mp \sqrt{a b})$, if $a>0, b>0, c>0$, or $(0 ; 0 ; 0),(\sqrt{b c} ; \pm \sqrt{a c} ; \mp \sqrt{a b})$, $(-\sqrt{b c} ; \pm \sqrt{a c} ; \pm \sqrt{a b})$, if $a<0, b<0, c<0$.