Is the cube of 4913 is an odd number? Why?
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Yes, It will be a odd no. because the cube root of an odd number is always odd
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![gien........ \\ \sqrt[3]{4913} \\ = (4913) { \frac{}{} }^{ \frac{1}{3} } \\ = (17 {}^{3} ) {}^{ \frac{1}{3} } \\ = 17 {}^{3 \times \frac{1}{3} } \\ = 17. \\ the \: result \: number \: \: is \: odd \: bcz \: \\ that \: is \: not \: multible \: of \: 2. \: if \: that \: divided \: by \: 2 \: result \: in \: fraction.](https://tex.z-dn.net/?f=gien........+%5C%5C++++%5Csqrt%5B3%5D%7B4913%7D++%5C%5C++%3D+%284913%29+%7B+%5Cfrac%7B%7D%7B%7D+%7D%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D++%5C%5C++%3D+%2817+%7B%7D%5E%7B3%7D+%29+%7B%7D%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D++%5C%5C++%3D+17+%7B%7D%5E%7B3+%5Ctimes++%5Cfrac%7B1%7D%7B3%7D+%7D++%5C%5C++%3D+17.+%5C%5C+the+%5C%3A+result+%5C%3A+number+%5C%3A++%5C%3A+is+%5C%3A+odd+%5C%3A+bcz+%5C%3A++%5C%5C+that+%5C%3A+is+%5C%3A+not+%5C%3A+multible+%5C%3A+of+%5C%3A+2.+%5C%3A+if+%5C%3A+that+%5C%3A+divided+%5C%3A+by+%5C%3A+2+%5C%3A+result+%5C%3A+in+%5C%3A+fraction. "gien........ \\ \sqrt[3]{4913} \\ = (4913) { \frac{}{} }^{ \frac{1}{3} } \\ = (17 {}^{3} ) {}^{ \frac{1}{3} } \\ = 17 {}^{3 \times \frac{1}{3} } \\ = 17. \\ the \: result \: number \: \: is \: odd \: bcz \: \\ that \: is \: not \: multible \: of \: 2. \: if \: that \: divided \: by \: 2 \: result \: in \: fraction.")
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