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Answer :
To determine which number in the monomial [tex]\(215 x^{18} y^3 z^{21}\)[/tex] needs to be changed to make it a perfect cube, we need to ensure that the coefficient and each of the variable exponents in the expression are multiples of 3.
Let's go through the components one by one:
1. Exponents of the Variables:
- For [tex]\(x^{18}\)[/tex], the exponent 18 is already a multiple of 3.
- For [tex]\(y^3\)[/tex], the exponent 3 is also a multiple of 3.
- For [tex]\(z^{21}\)[/tex], the exponent 21 is a multiple of 3 as well.
This means that the variable part of the monomial is already a perfect cube.
2. Coefficient:
- The numerical coefficient is 215. To check if 215 is a perfect cube, we would need its prime factorization to show that each prime factor's power is a multiple of 3.
The prime factorization of 215 is done as follows:
- 215 can be divided by 5 once, giving 43.
- 43 is a prime number and cannot be divided further by 5 or any other primes up to its value.
So, the factorization of 215 is [tex]\(5^1 \times 43^1\)[/tex]. For a number to be a perfect cube, all primes in its factorization should be raised to powers that are multiples of 3, which 215 doesn't satisfy.
Therefore, 215 is not a perfect cube, and to make the entire monomial a perfect cube, 215 needs to be changed.
Conclusion:
We found that the number in the monomial that needs to be changed to make it a perfect cube is 215. Adjusting 215 to the nearest perfect cube, which is 216, accomplishes the goal. Hence, the correct answer is:
215
Let's go through the components one by one:
1. Exponents of the Variables:
- For [tex]\(x^{18}\)[/tex], the exponent 18 is already a multiple of 3.
- For [tex]\(y^3\)[/tex], the exponent 3 is also a multiple of 3.
- For [tex]\(z^{21}\)[/tex], the exponent 21 is a multiple of 3 as well.
This means that the variable part of the monomial is already a perfect cube.
2. Coefficient:
- The numerical coefficient is 215. To check if 215 is a perfect cube, we would need its prime factorization to show that each prime factor's power is a multiple of 3.
The prime factorization of 215 is done as follows:
- 215 can be divided by 5 once, giving 43.
- 43 is a prime number and cannot be divided further by 5 or any other primes up to its value.
So, the factorization of 215 is [tex]\(5^1 \times 43^1\)[/tex]. For a number to be a perfect cube, all primes in its factorization should be raised to powers that are multiples of 3, which 215 doesn't satisfy.
Therefore, 215 is not a perfect cube, and to make the entire monomial a perfect cube, 215 needs to be changed.
Conclusion:
We found that the number in the monomial that needs to be changed to make it a perfect cube is 215. Adjusting 215 to the nearest perfect cube, which is 216, accomplishes the goal. Hence, the correct answer is:
215
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