Question

Simplify :

$\left(i\right)8^{\frac{4}{3}}+{25}^{\frac{3}{2}}-{\left(\frac{1}{27}\right)}^{-\frac{2}{3}}\left(ii\right)\left[\right(64{)}^{-2}{]}^{-3}÷\left[\right\{(-8{)}^2{\}}^3{]}^2(iii\left)\right(2^{-3}-2^{-4}\left)\right(2^{-3}+2^{-4})$

Answer 1

$\left(i\right)8^{\frac{4}{3}}+{25}^{\frac{3}{2}}-{\left(\frac{1}{27}\right)}^{-\frac{2}{3}}={\left(2^3\right)}^{\frac{4}{3}}+(5^2{)}^{\frac{3}{2}}-{\left(\frac{1}{3^3}\right)}^{-\frac{2}{3}}=2^{3\times \frac{4}{3}}+5^{2\times \frac{3}{2}}-\frac{1}{3^{3\times \left(\frac{-2}{3}\right)}}=2^4+5^3-\frac{1}{3^{-2}}=16+125-3^2=141-9=132(ii\left)\right[(64{)}^{-2}{]}^{-3}÷[\left\{\right(-8{)}^2{\}}^3{]}^2={\left(2^6\right)}^{-2\times -3}÷{(-8)}^{2\times 3\times 2}=2^{6\times \left(6\right)}÷{(-8)}^{12}=2^{+36}÷(-8{)}^{12}=2^{+36}÷[(-2{)}^3{]}^{12}=2^{36}÷(-2{)}^{36}=\frac{2^{36}}{(-2{)}^{36}}=\frac{2^{36}}{2^{36}}(\because 36\text{is even})=2^{36-36}=2^0=1(\because a^0=1)\left(iii\right)(2^{-3}-2^{-4})(2^{-3}+2^{-4})=(2^{-3}{)}^2-(2^{-4}{)}^2\{\because (a-b\left)\right(a+b)=a^2-b^2\}=2^{-6}-2^{-8}=\frac{1}{2^6}-\frac{1}{2^8}=\frac{1}{64}-\frac{1}{256}=\frac{4-1}{256}=\frac{3}{256}$