Question

Question 14
Simplify
$\left(i\right)(1^3+2^3+3^3{)}^{\frac{1}{2}}(ii\left){\left(\frac{3}{5}\right)}^4{\left(\frac{8}{5}\right)}^{-12}{\left(\frac{32}{5}\right)}^6\right(iii\left){\left(\frac{1}{27}\right)}^{\frac{-2}{3}}\right(iv\left){\left[{\left(\right(625{)}^{-\frac{1}{2}})}^{\frac{-1}{4}}\right]}^2\right(v)\frac{9^{\frac{1}{3}}\times {27}^{-\frac{1}{2}}}{3^{\frac{1}{6}}\times 3^{\frac{-2}{3}}}$
$\left(vi\right){64}^{\frac{-1}{3}}[{64}^{\frac{1}{3}}-{64}^{\frac{2}{3}}]\left(vii\right)\frac{8^{\frac{1}{3}}\times {16}^{\frac{1}{3}}}{{32}^{\frac{-1}{3}}}$

Answer 1

(i)
$(1^3+2^3+3^3{)}^{\frac{1}{2}}=(1+8+27{)}^{\frac{1}{2}}=(36{)}^{\frac{1}{2}}=(6^2{)}^{\frac{1}{2}}=6^{2\times \frac{1}{2}}=6[\because (a^m{)}^n=a^{mn}]$

(ii)
${\left(\frac{3}{5}\right)}^4{\left(\frac{8}{5}\right)}^{-12}{\left(\frac{32}{5}\right)}^6=\frac{3^4}{5^4}\times {\left(\frac{5}{2^3}\right)}^{12}{\left(\frac{2^5}{5}\right)}^6(\because a^{-1}=\frac{1}{a})$
$=\frac{3^4}{5^4}\times \frac{5^{12}}{2^{36}}\times \frac{2^{30}}{5^6}[\because (a^m{)}^n=a^{mn}]=\frac{3^4\times 5^{12-4-6}}{2^{36-30}}[\because \frac{a^m}{a^n}=a^{m-n}]=\frac{3^4}{2^6}\times 5^2=\frac{81\times 25}{64}=\frac{2025}{64}$

(iii) ${\left(\frac{1}{27}\right)}^{-2/3}={\left(\frac{1}{3^3}\right)}^{-2/3}[\because \frac{1}{a}=a^{-1}]=3^{-3\times \frac{-2}{3}}=3^2=9[\because (a^m{)}^n=a^{mn}]$

(iv)
${\left[{\left(\right(625{)}^{\frac{-1}{2}})}^{\frac{-1}{4}}\right]}^2={\left[{\left(\right(625{)}^{\frac{-1}{2}})}^{\frac{-1}{4}}\right]}^2=({25}^{-1}{)}^{\frac{-1}{4}\times 2}=[\left(5^2{)}^{-1}{]}^{\frac{-1}{4}\times 2}\right[\because \left(a^m{)}^n\right]=a^{mn}=5^{-2\times \frac{-1}{4}\times 2}=5^1=5$

(v)
$\frac{9^{\frac{1}{3}}\times {27}^{\frac{-1}{2}}}{3^{\frac{1}{6}}\times 3^{\frac{-2}{3}}}=\frac{(3^2{)}^{\frac{1}{3}}\times (3^3{)}^{\frac{-1}{2}}}{3^{\frac{1}{6}}\times 3^{\frac{-2}{3}}}[\because (a^m{)}^n=a^{mn}]$
$=\frac{3^{\frac{2}{3}}\times 3^{\frac{-3}{2}}}{3^{\frac{1}{6}}\times 3^{\frac{-2}{3}}}=\frac{3^{\frac{2}{3}-\frac{3}{2}}}{3^{\frac{1}{6}-\frac{2}{3}}}[\because a^m\times a^n=a^{m+n}]=\frac{3^{\frac{4-9}{6}}}{3^{\frac{1-4}{6}}}=\frac{3^{\frac{-5}{6}}}{3^{\frac{-3}{6}}}=3^{\frac{-5}{6}+\frac{3}{6}}=3^{\frac{-2}{6}}=3^{\frac{-1}{3}}[\because \frac{a^m}{a^n}=a^{m-n}]$

(vi)
${64}^{\frac{-1}{3}}[{64}^{\frac{1}{3}}-{64}^{\frac{2}{3}}]=(4^3{)}^{-\frac{1}{3}}\left[\right(4^3{)}^{\frac{1}{3}}-\left(4^3{)}^{\frac{2}{3}}\right]=4^{3\times \frac{-1}{3}}(4^{3\times \frac{1}{3}}-4^{3\times \frac{2}{3}})=4^{-1}(4-4^2\left)\right[\because (a^m{)}^n=a^{mn}]=\frac{1}{4}(4-16)=-\frac{12}{4}=-3$

(vii)
$\frac{8^{\frac{1}{3}}\times {16}^{\frac{1}{3}}}{{32}^{-\frac{1}{3}}}=\frac{(2^3{)}^{\frac{1}{3}}\times (2^4{)}^{\frac{1}{3}}}{(2^5{)}^{-\frac{1}{3}}}=\frac{2^{3\times \frac{1}{3}}\times 2^{4\times \frac{1}{3}}}{2^{5\times -\frac{1}{3}}}[\because (a^m{)}^n=a^{mn}]$
$=\frac{2^{\frac{3}{3}+\frac{4}{3}}}{2^{-\frac{5}{3}}}=\frac{2^{\frac{7}{3}}}{2^{-\frac{5}{3}}}=2^{\frac{7}{3}+\frac{5}{3}}=2^{\frac{12}{3}}=2^4=16[\because \frac{a^m}{a^n}=a^{m-n}]$