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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 analyze each part of the monomial. A perfect cube is a number that can be expressed as something raised to the power of 3.
The monomial contains four parts:
1. The coefficient [tex]\(215\)[/tex]
2. The exponent of [tex]\(x\)[/tex], which is [tex]\(18\)[/tex]
3. The exponent of [tex]\(y\)[/tex], which is [tex]\(3\)[/tex]
4. The exponent of [tex]\(z\)[/tex], which is [tex]\(21\)[/tex]
### 1. Coefficient: 215
Let's start with the coefficient [tex]\(215\)[/tex]. For [tex]\(215\)[/tex] to be a perfect cube, it must be able to be expressed as [tex]\(a^3\)[/tex] for some integer [tex]\(a\)[/tex].
We can check if [tex]\(215\)[/tex] is a perfect cube by checking its prime factorization. The prime factors of [tex]\(215\)[/tex] are:
[tex]\[ 215 = 5 \times 43 \][/tex]
Both 5 and 43 are prime numbers, and neither 5 nor 43 can be written as [tex]\(a^3\)[/tex] (they don't appear in groups of three). Therefore, [tex]\(215\)[/tex] is not a perfect cube.
### 2. Exponent of [tex]\(x\)[/tex]: 18
For the exponent [tex]\(18\)[/tex] to result in [tex]\(x\)[/tex] being a term in a perfect cube, it must be divisible by 3 because anything to the power of 3 involves exponents that are multiples of 3.
[tex]\[ 18 \div 3 = 6 \][/tex]
Since 18 is divisible by 3, the exponent [tex]\(18\)[/tex] does not need to be changed.
### 3. Exponent of [tex]\(y\)[/tex]: 3
For the exponent [tex]\(3\)[/tex] to result in [tex]\(y\)[/tex] being a term in a perfect cube, it also must be divisible by 3.
[tex]\[ 3 \div 3 = 1 \][/tex]
Since 3 is divisible by 3, the exponent [tex]\(3\)[/tex] does not need to be changed.
### 4. Exponent of [tex]\(z\)[/tex]: 21
For the exponent [tex]\(21\)[/tex] to result in [tex]\(z\)[/tex] being a term in a perfect cube, it too must be divisible by 3.
[tex]\[ 21 \div 3 = 7 \][/tex]
Since 21 is divisible by 3, the exponent [tex]\(21\)[/tex] does not need to be changed.
### Conclusion
From our analysis:
- The coefficient [tex]\(215\)[/tex] is not a perfect cube and needs to be changed to make the entire monomial a perfect cube.
- The exponents [tex]\(18\)[/tex], [tex]\(3\)[/tex], and [tex]\(21\)[/tex] are already multiples of 3, so they do not need to be changed.
Therefore, the number in the monomial [tex]\(215 x^{18} y^3 z^{21}\)[/tex] that needs to be changed to make it a perfect cube is:
[tex]\[ \boxed{215} \][/tex]
The monomial contains four parts:
1. The coefficient [tex]\(215\)[/tex]
2. The exponent of [tex]\(x\)[/tex], which is [tex]\(18\)[/tex]
3. The exponent of [tex]\(y\)[/tex], which is [tex]\(3\)[/tex]
4. The exponent of [tex]\(z\)[/tex], which is [tex]\(21\)[/tex]
### 1. Coefficient: 215
Let's start with the coefficient [tex]\(215\)[/tex]. For [tex]\(215\)[/tex] to be a perfect cube, it must be able to be expressed as [tex]\(a^3\)[/tex] for some integer [tex]\(a\)[/tex].
We can check if [tex]\(215\)[/tex] is a perfect cube by checking its prime factorization. The prime factors of [tex]\(215\)[/tex] are:
[tex]\[ 215 = 5 \times 43 \][/tex]
Both 5 and 43 are prime numbers, and neither 5 nor 43 can be written as [tex]\(a^3\)[/tex] (they don't appear in groups of three). Therefore, [tex]\(215\)[/tex] is not a perfect cube.
### 2. Exponent of [tex]\(x\)[/tex]: 18
For the exponent [tex]\(18\)[/tex] to result in [tex]\(x\)[/tex] being a term in a perfect cube, it must be divisible by 3 because anything to the power of 3 involves exponents that are multiples of 3.
[tex]\[ 18 \div 3 = 6 \][/tex]
Since 18 is divisible by 3, the exponent [tex]\(18\)[/tex] does not need to be changed.
### 3. Exponent of [tex]\(y\)[/tex]: 3
For the exponent [tex]\(3\)[/tex] to result in [tex]\(y\)[/tex] being a term in a perfect cube, it also must be divisible by 3.
[tex]\[ 3 \div 3 = 1 \][/tex]
Since 3 is divisible by 3, the exponent [tex]\(3\)[/tex] does not need to be changed.
### 4. Exponent of [tex]\(z\)[/tex]: 21
For the exponent [tex]\(21\)[/tex] to result in [tex]\(z\)[/tex] being a term in a perfect cube, it too must be divisible by 3.
[tex]\[ 21 \div 3 = 7 \][/tex]
Since 21 is divisible by 3, the exponent [tex]\(21\)[/tex] does not need to be changed.
### Conclusion
From our analysis:
- The coefficient [tex]\(215\)[/tex] is not a perfect cube and needs to be changed to make the entire monomial a perfect cube.
- The exponents [tex]\(18\)[/tex], [tex]\(3\)[/tex], and [tex]\(21\)[/tex] are already multiples of 3, so they do not need to be changed.
Therefore, the number in the monomial [tex]\(215 x^{18} y^3 z^{21}\)[/tex] that needs to be changed to make it a perfect cube is:
[tex]\[ \boxed{215} \][/tex]
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