Respuesta :

\left[a _{3}\right] = \left[ \frac{ - b^{2}}{6}+\frac{\frac{ - b^{4}}{3}+\left( \frac{-1}{3}\,i \right) \,\sqrt{3}\,b^{4}}{2^{\frac{2}{3}}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{24}+\left( \frac{1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{\sqrt[3]{2}}\right][a​3​​]=​⎣​⎢​⎢​⎢​⎢​⎡​​​6​​−b​2​​​​+​2​​3​​2​​​​​3​​√​(−1296−432b​2​​−16b​6​​+√​(1679616+1119744b​2​​+186624b​4​​+41472b​6​​+13824b​8​​)​​​)​​​​​​3​​−b​4​​​​+(​3​​−1​​i)√​3​​​b​4​​​​+​​3​​√​2​​​​​​24​​−​3​​√​(−1296−432b​2​​−16b​6​​+√​(1679616+1119744b​2​​+186624b​4​​+41472b​6​​+13824b​8​​)​​​)​​​​​+(​24​​1​​i)√​3​​​​3​​√​(−1296−432b​2​​−16b​6​​+√​(1679616+1119744b​2​​+186624b​4​​+41472b​6​​+13824b​8​​)​​​)​​​​​​⎦​⎥​⎥​⎥​⎥​⎤​​
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