QUESTION

# Simplify the given expression.${{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}$

Hint: Find the multiplicative inverse of the function. Undergo algebraic operations like subtraction and multiplication, thus simplify the given expression.

Given is the expression ${{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}$.
A multiplicative inverse or reciprocal for a number x is denoted by $\dfrac{1}{x}$ or ${{x}^{-1}}$ is a number which when multiplied by x yields the multiplicative identity 1.
The multiplicative inverse of a fraction $\dfrac{a}{b}$ is $\dfrac{b}{a}$. For the multiplicative inverse of a real number, divide 1 by the number.
We have been given, ${{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}..........(1)$
${{4}^{-1}}={\dfrac{1}{4}},{{5}^{-1}}={\dfrac{1}{5}},{{\left( \dfrac{5}{8} \right)}^{-1}}=\dfrac{8}{5}$, these are the multiplicative inverse of the given real number. Now let us substitute the values in equation (1).
${{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}={{\left( \dfrac{1}{4}-\dfrac{1}{5} \right)}^{2}}\times \left( \dfrac{8}{5} \right)$
\begin{align} & ={{\left( \dfrac{5-4}{4\times 5} \right)}^{2}}\times \left( \dfrac{8}{5} \right)={{\left( \dfrac{1}{20} \right)}^{2}}\times \dfrac{8}{5} \\ & =\dfrac{1}{20}\times \dfrac{1}{20}\times \dfrac{8}{5}=\dfrac{1}{20}\times \dfrac{2}{5}\times \dfrac{1}{5} \\ & =\dfrac{1}{10\times 5\times 5}=\dfrac{1}{10\times 25}=\dfrac{1}{250}=0.004 \\ \end{align}
${{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}=\dfrac{1}{250}=0.004$
$\therefore$We have simplified the given expression.