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 must ensure that all parts of the expression meet the criteria for being a perfect cube.
A perfect cube means any integer or variable is raised to a power that is a multiple of 3. Let's analyze each component:
1. Exponents of Variables:
- [tex]\(x^{18}\)[/tex]: The exponent is 18. Since [tex]\(18 \div 3 = 6\)[/tex], it is already a multiple of 3, so 18 is suitable for a perfect cube.
- [tex]\(y^3\)[/tex]: The exponent is 3. Since [tex]\(3 \div 3 = 1\)[/tex], it is also a multiple of 3, so 3 is already suitable.
- [tex]\(z^{21}\)[/tex]: The exponent is 21. Since [tex]\(21 \div 3 = 7\)[/tex], it is a multiple of 3, allowing 21 to be suitable for a perfect cube.
2. Coefficient:
- The coefficient is 215. For 215 to be a perfect cube, its prime factorization must involve prime factors each raised to a power that is a multiple of 3. Let's see the factorization:
- 215 can be factored into [tex]\(5 \times 43\)[/tex].
- Since 43 is a prime number and neither 5 nor 43 is raised to a power that is a multiple of 3, 215 is not a perfect cube.
In conclusion, while the exponents for [tex]\(x^{18}\)[/tex], [tex]\(y^3\)[/tex], and [tex]\(z^{21}\)[/tex] already meet the conditions for a perfect cube, the coefficient 215 does not. Therefore, the number that needs to be changed to make the entire expression a perfect cube is 215.
A perfect cube means any integer or variable is raised to a power that is a multiple of 3. Let's analyze each component:
1. Exponents of Variables:
- [tex]\(x^{18}\)[/tex]: The exponent is 18. Since [tex]\(18 \div 3 = 6\)[/tex], it is already a multiple of 3, so 18 is suitable for a perfect cube.
- [tex]\(y^3\)[/tex]: The exponent is 3. Since [tex]\(3 \div 3 = 1\)[/tex], it is also a multiple of 3, so 3 is already suitable.
- [tex]\(z^{21}\)[/tex]: The exponent is 21. Since [tex]\(21 \div 3 = 7\)[/tex], it is a multiple of 3, allowing 21 to be suitable for a perfect cube.
2. Coefficient:
- The coefficient is 215. For 215 to be a perfect cube, its prime factorization must involve prime factors each raised to a power that is a multiple of 3. Let's see the factorization:
- 215 can be factored into [tex]\(5 \times 43\)[/tex].
- Since 43 is a prime number and neither 5 nor 43 is raised to a power that is a multiple of 3, 215 is not a perfect cube.
In conclusion, while the exponents for [tex]\(x^{18}\)[/tex], [tex]\(y^3\)[/tex], and [tex]\(z^{21}\)[/tex] already meet the conditions for a perfect cube, the coefficient 215 does not. Therefore, the number that needs to be changed to make the entire expression a perfect cube is 215.
