Question

Prove that ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)=\frac{65}{16}$.

Answer 1

Step 1: Factorise each term into prime numbers.

${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)=\frac{65}{16}$

Take $\text{LHS}$

$\begin{array}{l}={\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)\\ ={\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)\\ ={\left(\frac{125}{64}\right)}^{\frac{2}{3}}+{\left(\frac{625}{256}\right)}^{\frac{1}{4}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)\left[\because p^{-n}=\frac{1}{p^n}\right]\\ ={\left(\frac{5\times 5\times 5}{4\times 4\times 4}\right)}^{\frac{2}{3}}+{\left(\frac{5\times 5\times 5\times 5}{4\times 4\times 4\times 4}\right)}^{\frac{1}{4}}+\left(\frac{\sqrt{5\times 5}}{\sqrt[3]{4\times 4\times 4}}\right)\\ ={\left(\frac{5^3}{4^3}\right)}^{\frac{2}{3}}+{\left(\frac{5^4}{4^4}\right)}^{\frac{1}{4}}+\left\{\frac{{\left(5^2\right)}^{\frac{1}{2}}}{{\left(4^3\right)}^{\frac{1}{3}}}\right\}\end{array}$

Step 2: Simplify the above expression using the law of radical exponents.

$\begin{array}{l}{\left(\frac{5^3}{4^3}\right)}^{\frac{2}{3}}+{\left(\frac{5^4}{4^4}\right)}^{\frac{1}{4}}+\left\{\frac{{\left(5^2\right)}^{\frac{1}{2}}}{{\left(4^3\right)}^{\frac{1}{3}}}\right\}\\ ={\left\{{\left(\frac{5}{4}\right)}^3\right\}}^{\frac{2}{3}}+{\left\{{\left(\frac{5}{4}\right)}^4\right\}}^{\frac{1}{4}}+\left\{\frac{{\left(5^2\right)}^{\frac{1}{2}}}{{\left(4^3\right)}^{\frac{1}{3}}}\right\}\\ ={\left(\frac{5}{4}\right)}^{3\times \frac{2}{3}}+{\left(\frac{5}{4}\right)}^{4\times \frac{1}{4}}+\left\{\frac{{\left(5\right)}^{2\times \frac{1}{2}}}{{\left(4\right)}^{3\times \frac{1}{3}}}\right\}\\ ={\left(\frac{5}{4}\right)}^2+{\left(\frac{5}{4}\right)}^1+\left(\frac{5}{4}\right)\\ =\frac{25}{16}+\frac{5}{4}+\frac{5}{4}\\ =\frac{25+20+20}{16}\\ =\frac{65}{16}=\text{RHS}\end{array}$

Hence it is proved that ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)=\frac{65}{16}$