[(4.9)^3+(2.1)^3]/[(4.9)^2-(10.29*1)+4.41]
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MathBot Answer:
\[\frac{{4.9}^{3} + {2.1}^{3}}{{4.9}^{2} - 10.29 \cdot 1 + 4.41} = 7\]
\[\begin{aligned}\frac{{4.9}^{3} + {2.1}^{3}}{{4.9}^{2} - 10.29 \cdot 1 + 4.41}& = \left(\left(4 + \frac{9}{10}\right)^{3} + \left(2 + \frac{1}{10}\right)^{3}\right) \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}}\\& = \left(\left(\frac{49}{10}\right)^{3} + \left(2 + \frac{1}{10}\right)^{3}\right) \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}}\\& = \left(\frac{117649}{1000} + \left(2 + \frac{1}{10}\right)^{3}\right) \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}}\\& = \left(\frac{117649}{1000} + \left(\frac{21}{10}\right)^{3}\right) \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}}\\& = \left(\frac{117649}{1000} + \frac{9261}{1000}\right) \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}}\\& = \frac{12691}{100} \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}}\\& = \frac{12691}{100} \cdot \frac{1}{\left(\frac{49}{10}\right)^{2} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}}\\& = \frac{12691}{100} \cdot \frac{1}{\frac{2401}{100} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}}\\& = \frac{12691}{100} \cdot \frac{1}{\frac{2401}{100} - \frac{1029}{100} \cdot 1 + 4 + \frac{41}{100}}\\& = \frac{12691}{100} \cdot \frac{1}{\frac{2401}{100} - \frac{1029}{100} + 4 + \frac{41}{100}}\\& = \frac{12691}{100} \cdot \frac{1}{\frac{343}{25} + 4 + \frac{41}{100}}\\& = \frac{12691}{100} \cdot \frac{1}{\frac{443}{25} + \frac{41}{100}}\\& = \frac{12691}{100} \cdot \frac{1}{\frac{1813}{100}}\\& = 7\end{aligned}\]