Question

Simplify \[{\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} \div {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}}\]

Answer 1

Hint: We write prime factorization of the number in the base and solve the term using the formula of exponents. Use the formula of exponents to cancel the terms within power so as to make the solution easier. Solve the two brackets and then divide the terms.

Formula used:
* Formula of exponents states that powers can be multiplied such that \[{({x^m})^{\dfrac{1}{n}}} = {x^{m \times \dfrac{1}{n}}} = {x^{\dfrac{m}{n}}}\]
* Prime factorization of a number is writing the number in multiples of its factors where all factors are prime numbers.
* When the base is same powers can be added i.e. \[{a^m} \times {a^n} = {a^{m + n}}\]
* If the power has negative sign, then we take reciprocal of the term i.e. \[{a^{ - 1}} = \dfrac{1}{a}\]

Complete step by step answer:
We have to find the value of \[{\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} \div {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}}\] ……………….… (1)
Since we know if the power has negative sign, then we take reciprocal of the term i.e. \[{a^{ - 1}} = \dfrac{1}{a}\]
\[ \Rightarrow {\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} \div {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}} = {\left( {216} \right)^{2/3}} \div {\left( {27} \right)^{4/3}}\] …………...… (2)
Now we solve the two terms in RHS of equation (2) separately
TERM 1:\[{\left( {216} \right)^{2/3}}\]
We write the prime factorization of the base
\[ \Rightarrow 216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\]
Since the bases are same we can add powers
\[ \Rightarrow 216 = {2^3} \times {3^3}\]
Since the powers are same we can multiply the bases
\[ \Rightarrow 216 = {(2 \times 3)^3}\]
\[ \Rightarrow 216 = {6^3}\]
Substitute the value of \[216 = {6^3}\] in the first term
\[ \Rightarrow {\left( {216} \right)^{2/3}} = {\left( {{6^3}} \right)^{2/3}}\]
Use the property of exponents i.e. \[{({x^m})^{\dfrac{1}{n}}} = {x^{m \times \dfrac{1}{n}}} = {x^{\dfrac{m}{n}}}\]
\[ \Rightarrow {\left( {216} \right)^{2/3}} = {(6)^{3 \times \dfrac{2}{3}}}\]
Cancel the same terms in numerator and denominator of the power
\[ \Rightarrow {\left( {216} \right)^{2/3}} = {(6)^2}\]
\[ \Rightarrow {\left( {216} \right)^{2/3}} = 2 \times 2 \times 3 \times 3\]
\[\therefore {\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} = 2 \times 2 \times 3 \times 3\] ………………….… (3)
TERM 2:\[{\left( {27} \right)^{ - 4/3}}\]
We write the prime factorization of the base
\[ \Rightarrow 27 = 3 \times 3 \times 3\]
Since the bases are same we can add powers
\[ \Rightarrow 27 = {3^3}\]
Substitute the value of \[27 = {3^3}\] in the first term
\[ \Rightarrow {\left( {27} \right)^{4/3}} = {\left( {{3^3}} \right)^{4/3}}\]
Use the property of exponents i.e. \[{({x^m})^{\dfrac{1}{n}}} = {x^{m \times \dfrac{1}{n}}} = {x^{\dfrac{m}{n}}}\]
\[ \Rightarrow {\left( {27} \right)^{4/3}} = {(3)^{3 \times \dfrac{4}{3}}}\]
Cancel the same terms in numerator and denominator of the power
\[ \Rightarrow {\left( {27} \right)^{4/3}} = {(3)^4}\]
\[ \Rightarrow {\left( {27} \right)^{4/3}} = 3 \times 3 \times 3 \times 3\]
\[\therefore {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}} = 3 \times 3 \times 3 \times 3\] ………………...… (4)
Substitute the values from equations (3) and (4) in equation (1)
\[ \Rightarrow {\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} \div {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}} = \left( {2 \times 2 \times 3 \times 3} \right) \div \left( {3 \times 3 \times 3 \times 3} \right)\]
Write the division in fraction form
\[ \Rightarrow {\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} \div {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}} = \dfrac{{\left( {2 \times 2 \times 3 \times 3} \right)}}{{\left( {3 \times 3 \times 3 \times 3} \right)}}\]
Cancel the same factors from both numerator and denominator
\[ \Rightarrow {\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} \div {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}} = \dfrac{{\left( {2 \times 2} \right)}}{{\left( {3 \times 3} \right)}}\]
\[ \Rightarrow {\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} \div {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}} = \dfrac{4}{9}\]

\[\therefore \] The value of \[{\left( {\dfrac{1}{{216}}} \right)^{ - 2/3}} \div {\left( {\dfrac{1}{{27}}} \right)^{ - 4/3}}\]is \[\dfrac{4}{9}\].

Note:
Students might make mistake of solving the power that exists in the power as they find it difficult to break the term \[ - \dfrac{2}{3}\]and \[ - \dfrac{4}{3}\] Keep in mind we should always write the term in the base in smallest and simplest form after collecting its powers so we can cancel as many terms in the power as possible.