15.若正数a、b满足$\frac{{a}^{2}}{{a}^{4}+{a}^{2}+1}$=$\frac{1}{24}$,$\frac{{b}^{3}}{{b}^{6}+{b}^{3}+1}$=$\frac{1}{19}$,则$\frac{ab}{({a}^{2}+a+1)({b}^{2}+b+1)}$=( )
0 287123 287131 287137 287141 287147 287149 287153 287159 287161 287167 287173 287177 287179 287183 287189 287191 287197 287201 287203 287207 287209 287213 287215 287217 287218 287219 287221 287222 287223 287225 287227 287231 287233 287237 287239 287243 287249 287251 287257 287261 287263 287267 287273 287279 287281 287287 287291 287293 287299 287303 287309 287317 366461
| A. | 24 | B. | 18 | C. | $\frac{1}{18}$ | D. | $\frac{1}{24}$ |