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 each part of the monomial can be expressed with exponents (or conditions) that are multiples of 3.
Here's how we can analyze this:
1. Coefficient (215):
- The coefficient should be a perfect cube. The prime factorization of [tex]\(215\)[/tex] is [tex]\(5 \times 43\)[/tex].
- None of the prime factors have exponents that are multiples of 3, indicating that 215 is not a perfect cube.
2. Exponent of [tex]\(x\)[/tex]:
- The exponent given is 18.
- 18 is divisible by 3 (since [tex]\(18 \div 3 = 6\)[/tex]), so it is already acceptable for a perfect cube.
3. Exponent of [tex]\(y\)[/tex]:
- The exponent given is 3.
- 3 is already divisible by 3 (since [tex]\(3 \div 3 = 1\)[/tex]), making it suitable for a perfect cube.
4. Exponent of [tex]\(z\)[/tex]:
- The exponent given is 21.
- 21 is divisible by 3 (since [tex]\(21 \div 3 = 7\)[/tex]), so it also satisfies the condition for a perfect cube.
Considering each component, the only part of the monomial that does not meet the condition for a perfect cube is the coefficient 215. If you want the monomial to be a perfect cube, you need to change the coefficient, as 215 is not itself a perfect cube. Therefore, the number that needs to be changed is [tex]\(215\)[/tex].
Here's how we can analyze this:
1. Coefficient (215):
- The coefficient should be a perfect cube. The prime factorization of [tex]\(215\)[/tex] is [tex]\(5 \times 43\)[/tex].
- None of the prime factors have exponents that are multiples of 3, indicating that 215 is not a perfect cube.
2. Exponent of [tex]\(x\)[/tex]:
- The exponent given is 18.
- 18 is divisible by 3 (since [tex]\(18 \div 3 = 6\)[/tex]), so it is already acceptable for a perfect cube.
3. Exponent of [tex]\(y\)[/tex]:
- The exponent given is 3.
- 3 is already divisible by 3 (since [tex]\(3 \div 3 = 1\)[/tex]), making it suitable for a perfect cube.
4. Exponent of [tex]\(z\)[/tex]:
- The exponent given is 21.
- 21 is divisible by 3 (since [tex]\(21 \div 3 = 7\)[/tex]), so it also satisfies the condition for a perfect cube.
Considering each component, the only part of the monomial that does not meet the condition for a perfect cube is the coefficient 215. If you want the monomial to be a perfect cube, you need to change the coefficient, as 215 is not itself a perfect cube. Therefore, the number that needs to be changed is [tex]\(215\)[/tex].
