Question

Find the value of

$\left(i\right){81}^{\frac{1}{3}}\times {576}^{\frac{1}{3}}$

$\left(ii\right){64}^{-\frac{2}{3}}\times {27}^{-\frac{2}{3}}$ [4 MARKS]

Answer 1

Application: 2 Marks each

$\left(i\right){81}^{\frac{1}{3}}\times {576}^{\frac{1}{3}}$

$=(81\times 576{)}^{\frac{1}{3}}$

$={\{(3^2\times 3^2)\times (4^3\times 3^2)\}}^{\frac{1}{3}}$

$={(3^{2+2+2}\times 4^3)}^{\frac{1}{3}}$

$=(3^6\times 4^3{)}^{\frac{1}{3}}$

$=(3^6{)}^{\frac{1}{3}}\times (4^3{)}^{\frac{1}{3}}$ $[Using:(x\times y{)}^a=x^a\times y^a]$

$=3^{6\times \frac{1}{3}}\times 4^{3\times \frac{1}{3}}$ $[Using:(x^a{)}^b=x^{ab}]$

$=3^2\times 4$

$=9\times 4$

$=36$

$\left(ii\right)(64{)}^{-\frac{2}{3}}\times (27{)}^{-\frac{2}{3}}$

$=(64\times 27{)}^{-\frac{2}{3}}$ $[Using:x^a\times y^a=(x\times y{)}^a]$

$=(4^3\times 3^3{)}^{-\frac{2}{3}}$

$={\left\{{(4\times 3)}^3\right\}}^{-\frac{2}{3}}$ $[\because a^n{b}^n=(ab{)}^n]$

$=({12}^3{)}^{-\frac{2}{3}}$

$={12}^{3\times -\frac{2}{3}}$ $[\because (x^a{)}^b=x^{ab}]$

$={12}^{-2}$

$=\frac{1}{{12}^2}$

$=\frac{1}{144}$