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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, let's break down the problem:
1. Understand Perfect Cubes:
- A perfect cube is a number that can be expressed as [tex]\(n^3\)[/tex] for some integer [tex]\(n\)[/tex].
- For a monomial to be a perfect cube, both the coefficient and each variable's exponent must be divisible by 3.
2. Evaluate the Coefficient: [tex]\(215\)[/tex]:
- The number [tex]\(215\)[/tex] needs to be checked to see if it can be a perfect cube. This involves checking if it can be expressed as a product of prime numbers, each raised to an exponent that is a multiple of 3.
- The prime factorization of [tex]\(215\)[/tex] is [tex]\(5 \times 43\)[/tex]. For [tex]\(215\)[/tex] to be a perfect cube, we need each of these primes to have exponents that are multiples of 3.
- Currently, [tex]\(5^1 \times 43^1\)[/tex] cannot be expressed as a perfect cube because neither exponent (1 in this case) is divisible by 3.
3. Evaluate the Exponents of Variables:
- [tex]\(x^{18}\)[/tex]: The exponent 18 is divisible by 3 ([tex]\(18 \div 3 = 6\)[/tex]). No change is needed.
- [tex]\(y^3\)[/tex]: The exponent 3 is divisible by 3 ([tex]\(3 \div 3 = 1\)[/tex]). No change is needed.
- [tex]\(z^{21}\)[/tex]: The exponent 21 is divisible by 3 ([tex]\(21 \div 3 = 7\)[/tex]). No change is needed.
4. Conclusion:
- Among the numbers in the monomial, the number 215 needs to be adjusted to make the entire expression a perfect cube. The primes [tex]\(5\)[/tex] and [tex]\(43\)[/tex] need to have exponents increased to multiples of 3 for the coefficient to be a perfect cube.
Therefore, the correct choice is to change [tex]\(215\)[/tex].
1. Understand Perfect Cubes:
- A perfect cube is a number that can be expressed as [tex]\(n^3\)[/tex] for some integer [tex]\(n\)[/tex].
- For a monomial to be a perfect cube, both the coefficient and each variable's exponent must be divisible by 3.
2. Evaluate the Coefficient: [tex]\(215\)[/tex]:
- The number [tex]\(215\)[/tex] needs to be checked to see if it can be a perfect cube. This involves checking if it can be expressed as a product of prime numbers, each raised to an exponent that is a multiple of 3.
- The prime factorization of [tex]\(215\)[/tex] is [tex]\(5 \times 43\)[/tex]. For [tex]\(215\)[/tex] to be a perfect cube, we need each of these primes to have exponents that are multiples of 3.
- Currently, [tex]\(5^1 \times 43^1\)[/tex] cannot be expressed as a perfect cube because neither exponent (1 in this case) is divisible by 3.
3. Evaluate the Exponents of Variables:
- [tex]\(x^{18}\)[/tex]: The exponent 18 is divisible by 3 ([tex]\(18 \div 3 = 6\)[/tex]). No change is needed.
- [tex]\(y^3\)[/tex]: The exponent 3 is divisible by 3 ([tex]\(3 \div 3 = 1\)[/tex]). No change is needed.
- [tex]\(z^{21}\)[/tex]: The exponent 21 is divisible by 3 ([tex]\(21 \div 3 = 7\)[/tex]). No change is needed.
4. Conclusion:
- Among the numbers in the monomial, the number 215 needs to be adjusted to make the entire expression a perfect cube. The primes [tex]\(5\)[/tex] and [tex]\(43\)[/tex] need to have exponents increased to multiples of 3 for the coefficient to be a perfect cube.
Therefore, the correct choice is to change [tex]\(215\)[/tex].
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