
Find the value of $\left[ {{{\left( {\dfrac{{ - 2}}{3}} \right)}^3} + \left( {\dfrac{4}{9}} \right)} \right] \div {\left( {\dfrac{5}{3}} \right)^3}$
Answer
536.7k+ views
Hint: - Use BODMAS rule.
$\left[ {{{\left( {\dfrac{{ - 2}}{3}} \right)}^3} + \left( {\dfrac{4}{9}} \right)} \right] \div {\left( {\dfrac{5}{3}} \right)^3}$
Simplify this equation using BODMAS rule
As we know ${\left( { - 2} \right)^3} = - 8,{\text{ }}{\left( 3 \right)^3} = 27,{\text{ and }}{\left( 5 \right)^3} = 125$
$
\Rightarrow \left[ {{{\left( {\dfrac{{ - 2}}{3}} \right)}^3} + \left( {\dfrac{4}{9}} \right)} \right] \div {\left( {\dfrac{5}{3}} \right)^3} \\
= \left[ {\dfrac{{ - 8}}{{27}} + \dfrac{4}{9}} \right] \div \dfrac{{125}}{{27}} \\
$
Multiply and divide by $3{\text{ in }}\dfrac{4}{9}$
$
= \left[ {\dfrac{{ - 8}}{{27}} + \left( {\dfrac{4}{9} \times \dfrac{3}{3}} \right)} \right] \div \dfrac{{125}}{{27}} \\
= \left[ {\dfrac{{ - 8}}{{27}} + \dfrac{{12}}{{27}}} \right] \div \dfrac{{125}}{{27}} \\
\Rightarrow \left[ {\dfrac{4}{{27}}} \right] \div \dfrac{{125}}{{27}} \\
$
Now above equation is written as
$\left[ {\dfrac{4}{{27}}} \right] \div \dfrac{{125}}{{27}} = \dfrac{{\dfrac{4}{{27}}}}{{\dfrac{{125}}{{27}}}} = \dfrac{4}{{27}} \times \dfrac{{27}}{{125}} = \dfrac{4}{{125}}$
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.
$\left[ {{{\left( {\dfrac{{ - 2}}{3}} \right)}^3} + \left( {\dfrac{4}{9}} \right)} \right] \div {\left( {\dfrac{5}{3}} \right)^3}$
Simplify this equation using BODMAS rule
As we know ${\left( { - 2} \right)^3} = - 8,{\text{ }}{\left( 3 \right)^3} = 27,{\text{ and }}{\left( 5 \right)^3} = 125$
$
\Rightarrow \left[ {{{\left( {\dfrac{{ - 2}}{3}} \right)}^3} + \left( {\dfrac{4}{9}} \right)} \right] \div {\left( {\dfrac{5}{3}} \right)^3} \\
= \left[ {\dfrac{{ - 8}}{{27}} + \dfrac{4}{9}} \right] \div \dfrac{{125}}{{27}} \\
$
Multiply and divide by $3{\text{ in }}\dfrac{4}{9}$
$
= \left[ {\dfrac{{ - 8}}{{27}} + \left( {\dfrac{4}{9} \times \dfrac{3}{3}} \right)} \right] \div \dfrac{{125}}{{27}} \\
= \left[ {\dfrac{{ - 8}}{{27}} + \dfrac{{12}}{{27}}} \right] \div \dfrac{{125}}{{27}} \\
\Rightarrow \left[ {\dfrac{4}{{27}}} \right] \div \dfrac{{125}}{{27}} \\
$
Now above equation is written as
$\left[ {\dfrac{4}{{27}}} \right] \div \dfrac{{125}}{{27}} = \dfrac{{\dfrac{4}{{27}}}}{{\dfrac{{125}}{{27}}}} = \dfrac{4}{{27}} \times \dfrac{{27}}{{125}} = \dfrac{4}{{125}}$
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.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Earth rotates from West to east ATrue BFalse class 6 social science CBSE

The easternmost longitude of India is A 97circ 25E class 6 social science CBSE

Write the given sentence in the passive voice Ann cant class 6 CBSE

Convert 1 foot into meters A030 meter B03048 meter-class-6-maths-CBSE

What is the LCM of 30 and 40 class 6 maths CBSE

Trending doubts
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE

How many crores make 10 million class 7 maths CBSE

AIM To prepare stained temporary mount of onion peel class 7 biology CBSE

The southernmost point of the Indian mainland is known class 7 social studies CBSE

Find HCF and LCM of 120 and 144 by using Fundamental class 7 maths CBSE

List of coprime numbers from 1 to 100 class 7 maths CBSE
