Question

Find the value of $\left[{\left({\displaystyle \frac{-2}{3}}\right)}^3+\left({\displaystyle \frac{4}{9}}\right)\right]÷{\left({\displaystyle \frac{5}{3}}\right)}^3$

Answer 1

Hint: - Use BODMAS rule.
$\left[{\left({\displaystyle \frac{-2}{3}}\right)}^3+\left({\displaystyle \frac{4}{9}}\right)\right]÷{\left({\displaystyle \frac{5}{3}}\right)}^3$
Simplify this equation using BODMAS rule
As we know ${\left(-2\right)}^3=-8,{\left(3\right)}^3=27,\text{ and }{\left(5\right)}^3=125$
$$\begin{gathered} \Rightarrow \left[{\left({\displaystyle \frac{-2}{3}}\right)}^3+\left({\displaystyle \frac{4}{9}}\right)\right]÷{\left({\displaystyle \frac{5}{3}}\right)}^3 \\ =\left[{\displaystyle \frac{-8}{27}}+{\displaystyle \frac{4}{9}}\right]÷{\displaystyle \frac{125}{27}} \end{gathered}$$
Multiply and divide by $3\text{ in }{\displaystyle \frac{4}{9}}$
$$\begin{gathered} =\left[{\displaystyle \frac{-8}{27}}+\left({\displaystyle \frac{4}{9}}\times {\displaystyle \frac{3}{3}}\right)\right]÷{\displaystyle \frac{125}{27}} \\ =\left[{\displaystyle \frac{-8}{27}}+{\displaystyle \frac{12}{27}}\right]÷{\displaystyle \frac{125}{27}} \\ \Rightarrow \left[{\displaystyle \frac{4}{27}}\right]÷{\displaystyle \frac{125}{27}} \end{gathered}$$
Now above equation is written as
$\left[{\displaystyle \frac{4}{27}}\right]÷{\displaystyle \frac{125}{27}}={\displaystyle \frac{{\displaystyle \frac{4}{27}}}{{\displaystyle \frac{125}{27}}}}={\displaystyle \frac{4}{27}}\times {\displaystyle \frac{27}{125}}={\displaystyle \frac{4}{125}}$
Note: - In such types of questions the key concept is that simplify this equation using BODMAS rule which is a bracket of division, multiplication, addition and subtraction, then after simplification we will get the required answer.