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# Is 128 a perfect cube? Find its cube root.

Last updated date: 13th Jun 2024
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Hint:
Here, we are required to find whether 128 is a perfect cube or not. We will first find the prime factors of the given number. Then, we will consider a group of triplets of the same numbers as a single number because we are required to find the cube root. If we are able to group each and every factor in any group of triplets then the given number is a perfect cube. Hence, multiplying the remaining factors, we would be able to find the required cube root of the given number.

Complete step by step solution:
To find whether a number is a perfect cube or not, we are required to use the prime factorization method.
We know that factorization is a method of writing an original number as the product of its various factors. Also, prime numbers are those numbers which are greater than 1 and have only two factors, i.e. factor 1 and the prime number itself.
Hence, prime factorization is a method in which we write the original number as the product of various prime numbers.
Therefore, prime factorization of 128 is:
$\begin{array}{*{20}{l}} 2\| \underline {128} \\ 2\| \underline {64} \\ 2\| \underline {32} \\ 2\| \underline {16} \\ 2\| \underline 8 \\ 2\| \underline 4 \\ 2\| \underline 2 \\ {}\| \underline 1 \end{array}$
Now, after doing the prime factorization, we are required to club the triplets of same prime factors together, as a single prime factor and if no factor is left out then the number is a perfect cube.
Hence, 128 can be written as:
$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Hence, we have expressed the given number as a product of its prime factors.
Now, we would group the triplets of same prime factors together, as a single prime factor:
$\Rightarrow 128 = \left( {2 \times 2 \times 2} \right) \times \left( {2 \times 2 \times 2} \right) \times 2$
Here, we are not able to group the factors without leaving behind any factor which is not in any group.
Hence, 128 is not a perfect cube.
Now, solving further, and converting the group of triplets as a single prime factor, we get,
$\sqrt[3]{{128}} = 2 \times 2 \times {2^{\dfrac{1}{3}}}$
$\Rightarrow \sqrt[3]{{128}} = 4\sqrt[3]{2}$
Hence, the Cube root of 128 is $4\sqrt[3]{2}$.

So, we can say that 128 is not a perfect cube and cube root of 128 is $4\sqrt[3]{2}$.

Note:
Here, we should know what a perfect cube means. If a number is multiplied thrice to itself then the number obtained is its cube. Cube root is the factor that is multiplied three times by itself to get the original number. The cube root of a perfect cube is always a whole number and not a fraction. If we get a fraction as a cube root then the number is not a perfect cube number.