Question

Find the value of \[{\left( {64} \right)^{ - \dfrac{2}{3}}}\]
a) \[16\]
b) \[\dfrac{1}{{16}}\]
c) \[ - \dfrac{1}{{16}}\]
d) None of these

Answer 1

Hint: The number resulted when a number is multiplied by itself is called a square of a number. A square of a number is an operation where a number is multiplied by itself. It is represented as \[{\left( x \right)^2} = x \times x\]
To find the cube root of a number by factorization, first, find the prime factors of the number and make a group of triplets of the same numbers from the prime factors and then find their products. For example, the prime factor of \[\left( c \right) = a \times a \times b \times b \times a \times b = \underline {\left[ {a \times a \times a} \right]} \times \underline {\left[ {b \times b \times b} \right]} = a \times b\]
In the question, we have to do both the operations of square and cube-root of a number for which factorize the number and find the power of the number and then multiply the power by the given power of the number.

Complete step by step answer:
Factorize the base of the expression by finding its prime numbers factors:
\[
   2\underline {\left| {64} \right.} \\
   2\underline {\left| {32} \right.} \\
   2\underline {\left| {16} \right.} \\
   2\underline {\left| 8 \right.} \\
   2\underline {\left| 4 \right.} \\
   2 \\
 \]
Hence the prime factor of 64 is:\[\left( {64} \right) = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]
This can also be written in factors power form as: \[\left( {64} \right) = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^6}\]
Now put the factorize power of 64 in the given expression as”
\[{\left( {64} \right)^{ - \dfrac{2}{3}}} = {\left( {{2^6}} \right)^{ - \dfrac{2}{3}}} = {\left( 2 \right)^{ - 6 \times \dfrac{2}{3}}}\]
Now solve the powers by cross multiplication:
\[{\left( 2 \right)^{ - 6 \times \dfrac{2}{3}}} = {\left( 2 \right)^{ - 2 \times 2}} = {\left( 2 \right)^{ - 4}}\]
We know \[{\left( x \right)^{ - n}}\] written as\[{\left( x \right)^{ - n}} = \dfrac{1}{{{x^n}}}\]hence ,
\[{\left( 2 \right)^{ - 4}} = \dfrac{1}{{{2^4}}} = \dfrac{1}{{16}}\] [Since\[{2^4} = 2 \times 2 \times 2 \times 2 = 16\]]
Hence, the value of \[{\left( {64} \right)^{ - \dfrac{2}{3}}} = \dfrac{1}{{16}}\]


Note: For any number which is to be reduced into powder form always try to bring the power of the number if the form of the power is given to it initially. It totally depends on the candidates which operation is either square or cube-root of the number they want to perform first. The result will remain the same irrespective of the process followed to solve the problem.