## Task A-1.4.

Let $a$ and $b$ be distinct real numbers and let $s=a-b$ and $t=a^{3}-b^{3}$.

Express $(a+b)^{2}$ in terms of $s$ and $t$.
## First Solution.

We can factorize $t=a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$, i.e., $t=s\left(a^{2}+a b+b^{2}\right)$.

After dividing by $s(s \neq 0)$, we get

$$
a^{2}+a b+b^{2}=\frac{t}{s}
$$

Cubing $s=a-b$ gives $s^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}=a^{3}-b^{3}-3 a b(a-b)=t-3 a b s$. Thus, $s^{3}=t-3 a b s$, from which we can easily express

$$
a b=\frac{t-s^{3}}{3 s}
$$

Finally, we get

$$
(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b=\frac{t}{s}+\frac{t-s^{3}}{3 s}=\frac{4 t-s^{3}}{3 s}
$$