15. If the lengths of the three sides of a triangle are $\mathrm{a}, \mathrm{b}$, $\mathrm{c}$, and $a^{3}+\frac{\sqrt{2}}{4} b^{3}+\frac{\sqrt{3}}{9} c^{3}-\frac{\sqrt{6}}{2} a b c=0$, then this triangle must be ( )
(A) Isosceles triangle
(B) Right triangle
(C) Equilateral triangle
(D) Isosceles right triangle
Answer: B

Solution:
$$
\begin{array}{c}
\text { Since } a^{3}+\frac{\sqrt{2}}{4} b^{3}+\frac{\sqrt{3}}{9} c^{3}-\frac{\sqrt{6}}{2} a b c=a^{3}+\left(\frac{\sqrt{2}}{2} b\right)^{3}+\left(\frac{\sqrt{3}}{3} c\right)^{3}-3 \cdot a \cdot \frac{\sqrt{2}}{2} b \cdot \frac{\sqrt{3}}{3} c=0, \\
\left(a+\frac{\sqrt{2}}{2} b+\frac{\sqrt{3}}{3} c\right) \cdot \frac{1}{2}\left[\left(a-\frac{\sqrt{2}}{2} b\right)^{2}+\left(a-\frac{\sqrt{3}}{3} c\right)^{2}+\left(\frac{\sqrt{2}}{2} b-\frac{\sqrt{3}}{3} c\right)^{2}\right]=0,
\end{array}
$$

Therefore, $a+\frac{\sqrt{2}}{2} b+\frac{\sqrt{3}}{3} c=0$ (discard) or $a=\frac{\sqrt{2}}{2} b=\frac{\sqrt{3}}{3} c$,

we get
$$
b=\sqrt{2} a \text { and } c=\sqrt{3} a \text {, }
$$

thus
$$
a^{2}+b^{2}=c^{2},
$$

so this triangle must be a right triangle.
Hence, the answer is (B).