3. The diagonals of a rhombus differ by $14 \mathrm{~cm}$. If we extend the longer diagonal by $2 \mathrm{~cm}$, and the shorter one by $8 \mathrm{~cm}$, the area of the rhombus will increase by $144 \mathrm{~cm}^{2}$. What are the lengths of the diagonals of the original rhombus?
First method:

Let $e_{1}$ and $f_{1}$ be the lengths of the diagonals of the original rhombus, and $e_{2}$ and $f_{2}$ be the lengths of the diagonals of the rhombus obtained after extending the diagonals. Then we have:

$e_{1}=f_{1}+14$,

$e_{2}=e_{1}+2=f_{1}+14+2=f_{1}+16 \text{ and}$

$f_{2}=f_{1}+8$.

It holds that $P_{2}=P_{1}+144$, or $\frac{e_{2} \cdot f_{2}}{2}=\frac{e_{1} \cdot f_{1}}{2}+144$.

Further, $\frac{\left(f_{1}+16\right) \cdot\left(f_{1}+8\right)}{2}=\frac{\left(f_{1}+14\right) \cdot f_{1}}{2}+144$,

$\left(f_{1}+16\right) \cdot\left(f_{1}+8\right)=\left(f_{1}+14\right) \cdot f_{1}+288$,

$f_{1}^{2}+16 f_{1}+8 f_{1}+128=f_{1}^{2}+14 f_{1}+288$,

$10 f_{1}=160 \Rightarrow f_{1}=16$.

Then $e_{1}=16+14=30$.

The lengths of the diagonals of the rhombus are $30 \text{ cm}$ and $16 \text{ cm}$.