## Task B-1.2.

If $a-b=3$, and $a \cdot b=1$, what is $\frac{1}{a^{3}}-\frac{1}{b^{3}}$ and $\frac{1}{a^{6}}+\frac{1}{b^{6}}$?
## Solution.

By squaring the equation $a-b=3$, we obtain

$$
a^{2}-2 a b+b^{2}=9
$$

from which it follows that

$$
a^{2}+b^{2}=9+2 a b=11
$$

Now we have

$$
\begin{array}{cc}
\frac{1}{a^{3}}-\frac{1}{b^{3}}=\frac{b^{3}-a^{3}}{(a b)^{3}}=\frac{(b-a)\left(b^{2}+a b+a^{2}\right)}{(a b)^{3}}=\frac{-3(11+1)}{1^{3}}=-36 & 2 \text { points } \\
\frac{1}{a^{6}}+\frac{1}{b^{6}}=\left(\frac{1}{a^{3}}-\frac{1}{b^{3}}\right)^{2}+2 \cdot \frac{1}{(a b)^{3}}=(-36)^{2}+2=1298 . & 2 \text { points }
\end{array}
$$