85(1195). The lengths of the sides of triangle $ABC$ are $a$ cm, $b$ cm, and $c$ cm. Knowing that $a^{3}=b^{3}+c^{3}$, determine whether angle $A$ is acute, right, or obtuse.

200
Solution. We will use the Law of Cosines, as the sign of the cosine can answer the question of the problem. We have:

$$
a^{2}=b^{2}+c^{2}-2 b c \cdot \cos A
$$

from which

$$
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}
$$

To use the given data ($a^{3}=b^{3}+c^{3}$), we multiply the numerator and the denominator of the fraction on the right side of equation (1) by $a(a \neq 0)$. We will have:

$$
\cos A=\frac{b^{2} a+c^{2} a-a^{3}}{2 a b c}=\frac{b^{2} a+c^{2} a-b^{3}-c^{3}}{2 a b c}=\frac{b^{2}(a-b)+c^{2}(a-c)}{2 a b c}
$$

Since $a^{3}=b^{3}+c^{3}$, then $a>b$ and $a>c$. Therefore, the numerator and the denominator of the fraction on the right side of equation (2) are positive, i.e., $\cos A>0$ and angle $A$ is acute.

Answer: angle $A$ is acute.