
How many numbers from the following are the perfect cubes?
(i) 125
(ii) 243
(iii) 343
(iv) 256
(v) 729
(vi) 1331
(vii) 8000
(viii) 9261
(ix) 5324
(x) 3375
Answer
612k+ views
Hint: Write the given number in prime factors and see if they are forming triplets of equal numbers like $2\times 2\times 2;\text{ }3\times 3\times 3;\text{ }5\times 5\times 5$ etc. then observe if any factor is left out or not. If no factor is left out then the number is a perfect cube.
Complete step-by-step answer:
A perfect cube is a number that is a cube of an integer. For example: 1, 8, 27, 64 are some examples of perfect cubes. In mathematics, a cube root of a number $x$ is a number $y$ such that ${{y}^{3}}=x$. All non-zero real numbers have exactly one real cube root. The most straightforward method to check if a number is a perfect cube or not is to check through factorization. If the prime factors of a number are grouped in triples of equal factors and no factor is left out then that number is a perfect cube.
Now, we come to the question. First we will find out the prime factors of the given numbers one by one and then try to group them in triplets.
(i) $125=5\times 5\times 5={{5}^{3}}$
(ii) \[243=3\times 3\times 3\times 3\times 3={{3}^{2}}\times {{3}^{3}}\]
(iii) $343=7\times 7\times 7={{7}^{3}}$
(iv) \[256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2={{2}^{2}}\times {{2}^{3}}\times {{2}^{3}}\]
(v) $729=3\times 3\times 3\times 3\times 3\times 3={{3}^{3}}\times {{3}^{3}}={{9}^{3}}$
(vi) $1331=11\times 11\times 11={{11}^{3}}$
(vii) $8000=8\times 1000={{2}^{3}}\times {{10}^{3}}={{20}^{3}}$
(viii) $9261=3\times 3\times 3\times 7\times 7\times 7={{3}^{3}}\times {{7}^{3}}$
(ix) $5324=2\times 2\times 11\times 11\times 11={{2}^{2}}\times {{11}^{2}}$
(x) $3375=3\times 3\times 3\times 5\times 5\times 5={{3}^{3}}\times {{5}^{3}}={{15}^{3}}$
From observing the above situations, we can conclude that: 125, 343, 729, 1331, 8000, 9261, 3375 are perfect cubes.
Notes: If there are groups of doublets then it will form a perfect square like 256 and 5324. Now, if we have to find the cube root then simply remove the exponent ‘3’ from the numbers which are perfect cubes. For example: 15 will be the cube root of 3375.
Complete step-by-step answer:
A perfect cube is a number that is a cube of an integer. For example: 1, 8, 27, 64 are some examples of perfect cubes. In mathematics, a cube root of a number $x$ is a number $y$ such that ${{y}^{3}}=x$. All non-zero real numbers have exactly one real cube root. The most straightforward method to check if a number is a perfect cube or not is to check through factorization. If the prime factors of a number are grouped in triples of equal factors and no factor is left out then that number is a perfect cube.
Now, we come to the question. First we will find out the prime factors of the given numbers one by one and then try to group them in triplets.
(i) $125=5\times 5\times 5={{5}^{3}}$
(ii) \[243=3\times 3\times 3\times 3\times 3={{3}^{2}}\times {{3}^{3}}\]
(iii) $343=7\times 7\times 7={{7}^{3}}$
(iv) \[256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2={{2}^{2}}\times {{2}^{3}}\times {{2}^{3}}\]
(v) $729=3\times 3\times 3\times 3\times 3\times 3={{3}^{3}}\times {{3}^{3}}={{9}^{3}}$
(vi) $1331=11\times 11\times 11={{11}^{3}}$
(vii) $8000=8\times 1000={{2}^{3}}\times {{10}^{3}}={{20}^{3}}$
(viii) $9261=3\times 3\times 3\times 7\times 7\times 7={{3}^{3}}\times {{7}^{3}}$
(ix) $5324=2\times 2\times 11\times 11\times 11={{2}^{2}}\times {{11}^{2}}$
(x) $3375=3\times 3\times 3\times 5\times 5\times 5={{3}^{3}}\times {{5}^{3}}={{15}^{3}}$
From observing the above situations, we can conclude that: 125, 343, 729, 1331, 8000, 9261, 3375 are perfect cubes.
Notes: If there are groups of doublets then it will form a perfect square like 256 and 5324. Now, if we have to find the cube root then simply remove the exponent ‘3’ from the numbers which are perfect cubes. For example: 15 will be the cube root of 3375.
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