2.290. Excluding $u$ and $v$ from the equations $u-v=a, u^{2}-v^{2}=b$, $u^{3}-v^{3}=c$, find the relation between $a, b$ and $c$.
## Solution.

$$
\left\{\begin{array} { l } 
{ u - v = a , } \\
{ u ^ { 2 } - v ^ { 2 } = b , } \\
{ u ^ { 3 } - v ^ { 3 } = c , }
\end{array} \left\{\begin{array}{l}
u-v=a \\
(u-v)(u+v)=b \\
(u-v)\left((u+v)^{2}-u v\right)=c
\end{array}\right.\right.
$$

Substituting $u-v=a$ from (1) into (2), we get $a(u+v)=b$, hence $u+v=\frac{b}{a}\left(2^{*}\right)$. Adding (1) and (2*), we have

$$
\left\{\begin{array}{l}
u-v=a \\
+ \\
\frac{u+v=\frac{b}{a}}{2 u=a+\frac{b}{a}}
\end{array}, \text { from which } u=\frac{a^{2}+b}{2 a}\right. \text { (4) }
$$

Subtracting (1) from (2*), we get

$$
\left\{\begin{array}{l}
u+v=\frac{b}{a} \\
- \\
\frac{u-v=a}{2 u=\frac{b}{a}-a}
\end{array}, \text { from which } v=\frac{b-a^{2}}{2 a}\right.
$$

Multiplying (4) by (5), we get $u v=\frac{a^{2}+b}{2 a} \cdot \frac{b-a^{2}}{2 a}=-\frac{a^{4}-b^{2}}{4 a^{2}}$

(6). Substituting $u-v=a$ from (1), $u+v=\frac{b}{a}$ from (2*), and $u v=-\frac{a^{4}-b^{2}}{4 a^{2}}$ from (6) into (3), we find $a\left(\frac{b^{2}}{a^{2}}+\frac{a^{4}-b^{2}}{4 a^{2}}\right)=c$, or $\frac{b^{2}}{a}+\frac{a^{4}-b^{2}}{4 a}=c$, or $4 b^{2}+a^{4}-b^{2}=4 a c$, or $3 b^{2}+a^{4}=4 a c$. Answer: $3 b^{2}+a^{4}=4 a c$.

Verify the validity of the equalities (2.291 - 2.304):