\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][a3]=⎣⎢⎢⎢⎢⎡6−b2+2323√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))3−b4+(3−1i)√3b4+3√224−3√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))+(241i)√33√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))⎦⎥⎥⎥⎥⎤
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Answer:
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