Answer
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Hint: In order to find the number to divide so that the number becomes a perfect cube, we should know what a perfect cube is. A perfect cube is a number which is equal to a number multiplied by itself three times. For example, if $a$ is a perfect cube of $b$, that means $a = {b^3}$. For the cube root of a perfect cube, we should get a natural number. Hence, $\sqrt[3]{{{b^3}}} = a$.
Complete step by step solution:
We are given a number $3087$.
Expanding this number in terms of its prime factors, and we get:
$3087 = 3 \times 3 \times 7 \times 7 \times 7$
We can see that there are two $3's$ and three $7's$.
For a perfect cube, we need multiplication of three same numbers, whose cube root return’s the same number.
So, here in the factors above there are only multiplications of three $7's$, that can give a perfect cube, because two $3's$ cannot form a cube.
If we can remove the two $3's$ from the factors of $3087$ , it can also give a perfect cube. And, it can be removed by dividing it by that number.
So, dividing $3087$ by $3 \times 3$, we get:
$\dfrac{{3087}}{{3 \times 3}} = \dfrac{{3 \times 3 \times 7 \times 7 \times 7}}{{3 \times 3}}$
On solving the terms, it gives:
$343 = 7 \times 7 \times 7$
And, we obtain a perfect cube, as $7 \times 7 \times 7 = 343$ and $\sqrt[3]{{343}} = 7$.
Therefore, the smallest number which can be divided from $3087$ to get the quotient as a perfect cube is $3 \times 3 = 9$.
Note:
If instead of dividing, there was multiplication that which smallest number can be multiplied to give a perfect cube, then we would have multiplied $3$ in order to make three $3's$, so that it completes one other perfect cube, and the whole number becomes a perfect cube.
Complete step by step solution:
We are given a number $3087$.
Expanding this number in terms of its prime factors, and we get:
$3087 = 3 \times 3 \times 7 \times 7 \times 7$
We can see that there are two $3's$ and three $7's$.
For a perfect cube, we need multiplication of three same numbers, whose cube root return’s the same number.
So, here in the factors above there are only multiplications of three $7's$, that can give a perfect cube, because two $3's$ cannot form a cube.
If we can remove the two $3's$ from the factors of $3087$ , it can also give a perfect cube. And, it can be removed by dividing it by that number.
So, dividing $3087$ by $3 \times 3$, we get:
$\dfrac{{3087}}{{3 \times 3}} = \dfrac{{3 \times 3 \times 7 \times 7 \times 7}}{{3 \times 3}}$
On solving the terms, it gives:
$343 = 7 \times 7 \times 7$
And, we obtain a perfect cube, as $7 \times 7 \times 7 = 343$ and $\sqrt[3]{{343}} = 7$.
Therefore, the smallest number which can be divided from $3087$ to get the quotient as a perfect cube is $3 \times 3 = 9$.
Note:
If instead of dividing, there was multiplication that which smallest number can be multiplied to give a perfect cube, then we would have multiplied $3$ in order to make three $3's$, so that it completes one other perfect cube, and the whole number becomes a perfect cube.
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