Question

Find the cubes of:

(i) $\frac{3}{7}$ (ii) $\frac{8}{9}$ (iii) $\frac{10}{13}$

(iv) $1\frac{2}{7}$ (v) $2\frac{1}{2}$

Answer 1

$\left(i\right)\frac{3}{7}={\left(\frac{3}{7}\right)}^3=\frac{3\times 3\times 3}{7\times 7\times 7}=\frac{27}{343}\left(ii\right)\frac{8}{9}={\left(\frac{8}{9}\right)}^3=\frac{8\times 8\times 8}{9\times 9\times 9}=\frac{512}{729}\left(iii\right)\frac{10}{13}={\left(\frac{10}{13}\right)}^3=\frac{10\times 10\times 10}{13\times 13\times 13}=\frac{1000}{2197}\left(iv\right)1\frac{2}{7}={\left(1\frac{2}{7}\right)}^3={\left(\frac{1\times 7+2}{7}\right)}^3={\left(\frac{9}{7}\right)}^3=\frac{9\times 9\times 9}{7\times 7\times 7}=\frac{729}{343}=2\frac{43}{343}\left(v\right)2\frac{1}{2}={\left(2\frac{1}{2}\right)}^3={\left(\frac{5}{2}\right)}^3=\frac{5\times 5\times 5}{2\times 2\times 2}=\frac{125}{8}=15\frac{5}{8}$