Question

Solve this:
Q.18. Simplify :
(a) $\frac{{\left(1^3+2^3+3^3\right)}^{{\displaystyle \frac{1}{2}}}}{{\left(4^3+{15}^2\right)}^{{\displaystyle \frac{1}{2}}}}$

Answer 1

Dear student,

$$\begin{gathered} a)\frac{{\left(1^3+2^3+3^3\right)}^{{\displaystyle \frac{1}{2}}}}{{\left(4^3+{15}^2\right)}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{{\displaystyle {\left(1+8+27\right)}^{\frac{1}{2}}}}{{{\displaystyle \left(64+225\right)}}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{{\displaystyle {\left(36\right)}^{\frac{1}{2}}}}{{\displaystyle {\left(289\right)}^{\frac{1}{2}}}} \\ =\frac{{\left(6^2\right)}^{{\displaystyle \frac{1}{2}}}}{{\left({17}^2\right)}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{{\left(6\right)}^{2\times {\displaystyle \frac{1}{2}}}}{{\left(17\right)}^{2\times {\displaystyle \frac{1}{2}}}}--[u\sin g{\left(a^m\right)}^n=a^{m\times n}] \\ =\frac{6}{17} \end{gathered}$$

Regards