Question

Is 9261 a perfect cube? In case of a perfect cube, find the number whose cube it is.

Answer 1

Hint:
1) A cube is a number which is a product of three identical numbers. e.g. $125=5\times 5\times 5$ is a cube of 5. In other words, it is the third power of a number. The cube is also the number multiplied by its square: e.g. ${{5}^{3}}={{5}^{2}}\times 5$ .
2) A perfect cube means the cube of an integer.
3) It can be noted that ${{20}^{3}}=8000$ and ${{30}^{3}}=27000$ . So, 9261 must be a cube of some number between 20 and 30, in fact, of a number which is very close to 20.
4) The powers of all prime numbers in the prime-factorization of a cube are multiples of 3.

Complete step by step solution:
Let us factorize 9261 into prime factors to determine whether it is a cube or not.
It is easy to see that 9261 is not divisible by 2 or 5 and that it is a multiple of 3. So, let's divide 9261 by 3.
 $9261= 3\times 3087$
3087 can be divided by 3 again.
 $9261=3\times 3\times 1029$
By dividing repeatedly by prime numbers, we finally get the following factorization:
 $9261=3\times 3\times 3\times 7\times 7\times 7$
Which can be written as:
 $9261={{3}^{3}}\times {{7}^{3}}$
So, 9261 is a perfect cube because all the powers of the prime numbers are multiples of 3.
Also, $9261={{3}^{3}}\times {{7}^{3}}={{\left( 3\times 7 \right)}^{3}}={{21}^{3}}$.

Therefore, 9261 is the cube of 21.

Note:
The last digit of a perfect cube can be any digit from 0 to 9.
If $x={{y}^{3}}$ , then we say that "x is the cube of y" and "y is the cube-root of x (denoted by $y={{x}^{\tfrac{1}{3}}}=\sqrt[3]{x}$ )".
Cubes of positive numbers are positive and cubes of negative numbers are negative. Generally speaking, even powers are always positive, whereas odd powers of negative numbers are negative.