Question

How do you simplify \[{\left( {\dfrac{{64}}{{125}}} \right)^{\dfrac{1}{3}}}?\]

Answer 1

Hint: We will factorise the terms in numerator and denominator and, then will try to find the single factor for cubic root from the numerator and denominator. After doing some simplification we get the required answer.

Formula used: If any variable is given in the form of a cube, it means that it has the same number or same digits or same prime factors multiplied three times in a row.
Let consider an example:
\[V\] is a variable and it is equivalent to another prime factor of \[a\].
It states that: \[V = {a^3}\].
Now we can rewrite as: \[V = a \times a \times a\]
Similarly, if we try to find the cubic root of any number, it simply states that we have to find the three same kinds of prime factors in that number.
Let say, a number \[{V_2}\]is represented as \[{V_2} = \sqrt[3]{a}\], where \[a\] is a number.
Then \[a\] must has three prime factors or \[a\]can be written as multiplication of three same prime factors
Or, it can be represented as: \[{V_2} = \sqrt[3]{{b \times b \times b}}\], where \[a = b \times b \times b\].
Then only, we can rewrite it as: \[{V_2} = b\].

Complete step-by-step solution:
Now imagine, if any number is given in cubic form of its reciprocal, then also we can split it up into the multiplication of three different prime factors or numbers.
Likewise;
Let say, \[{V_1}\] is a variable, which denotes that \[{V_1} = {\left( {\dfrac{1}{b}} \right)^3}\], where \[b\] is a prime factor.
So, in this case we can also rewrite the above iteration in following form:
\[{V_1} = \dfrac{1}{b} \times \dfrac{1}{b} \times \dfrac{1}{b}\].
So, if any number is given under a cubic root in its reciprocal form, then that number has to have three prime factors of it.
Let say, \[{V_3}\] is a number and, it is represented as: \[{V_3} = \sqrt[3]{{\dfrac{1}{{{a^3}}}}}\], where \[a\] is a prime factor.
Then we can simply write it as \[{V_3} = \dfrac{1}{a}\] .
Now, in the question, we have to simplify\[{\left( {\dfrac{{64}}{{125}}} \right)^{\dfrac{1}{3}}}\].
So, we will try to factorize the numerator and denominator separately and, then we will try to implement them into cubic root form.
Let say, \[P = {\left( {\dfrac{{64}}{{125}}} \right)^{\dfrac{1}{3}}}\].
Now, if we factorize \[64\] into its prime factors then we can write the following iteration:
\[64 = (2 \times 2 \times 2 \times 2 \times 2 \times 2)\].
Now, we have to find a triplet of a prime factor so that we can find the cubic root easily from it.
So, We can write it as: \[64 = (2 \times 2 \times 2) \times (2 \times 2 \times 2)\]
Or, we can rewrite it as: \[64 = {2^3} \times {2^3}\].
Again, in a similar way, we can write \[125\].
\[125 = 5 \times 5 \times 5\].
Or, \[125 = {5^3}\].
So, we can split \[P\] as following form:
\[P = {\left( {64 \times \dfrac{1}{{125}}} \right)^{\dfrac{1}{3}}}\]
By putting the above values:
\[P = {\left( {{2^3} \times {2^3} \times \dfrac{1}{{{5^3}}}} \right)^{\dfrac{1}{3}}}\]
So, it is clear that the prime number \[2\] and \[5\] has a triplet under the cubic root.
So, we can write the following equation:
\[P = \left( {2 \times 2 \times \dfrac{1}{5}} \right)\]
Or, \[P = \left( {\dfrac{4}{5}} \right)\]
After division, we get:
\[P = 0.8\].

\[\therefore \] The value of \[{\left( {\dfrac{{64}}{{125}}} \right)^{\dfrac{1}{3}}}\] is \[0.8\] or \[\dfrac{4}{5}\]

Note: Alternate process:
We can also use the general rule of indices as the number under the cubic root has to have a triplet.
So, we can solve the above equation as:
\[P = {\left( {\dfrac{{64}}{{125}}} \right)^{\dfrac{1}{3}}}\]
Or, \[P = {\left( {{2^3} \times {2^3} \times \dfrac{1}{{{5^3}}}} \right)^{\dfrac{1}{3}}}\]
Now using the general rule of indices:
\[P = \left( {{2^{3 \times \dfrac{1}{3}}} \times {2^{3 \times \dfrac{1}{3}}} \times \dfrac{1}{{{5^{3 \times \dfrac{1}{3}}}}}} \right)\]
So, \[P = \left( {{2^1} \times {2^1} \times \dfrac{1}{{{5^1}}}} \right)\]
Or, \[P = \dfrac{4}{5}\].
Always remember that:
Cube of a number is denoted in \[{()^3}\] form.
Cubic root of any number is denoted in \[{()^{\dfrac{1}{3}}}\] or \[\sqrt[3]{{()}}\] form.