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 look at the exponents of each term and the numerical coefficient.
1. Examine the exponents: For a monomial to be a perfect cube, all exponents in the expression, including the coefficient, must be multiples of 3.
- The exponent of [tex]\(x\)[/tex] is 18. Since 18 is a multiple of 3 (because [tex]\(18 \div 3 = 6\)[/tex]), [tex]\(x^{18}\)[/tex] is already a perfect cube.
- The exponent of [tex]\(y\)[/tex] is 3. Since 3 is a multiple of 3, [tex]\(y^3\)[/tex] is already a perfect cube.
- The exponent of [tex]\(z\)[/tex] is 21. Since 21 is a multiple of 3 (because [tex]\(21 \div 3 = 7\)[/tex]), [tex]\(z^{21}\)[/tex] is already a perfect cube.
2. Examine the coefficient (215): For the entire monomial to be a perfect cube, 215 must itself be a perfect cube. Let's consider its prime factorization:
- The prime factorization of 215 is [tex]\(5 \times 43\)[/tex].
A number is a perfect cube if each prime factor's exponent in the factorization is a multiple of 3. Here, [tex]\(5^1\)[/tex] and [tex]\(43^1\)[/tex] both have exponents of 1, which are not multiples of 3. Therefore, 215 is not a perfect cube.
Since the exponents of the variables [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] are already suitable for a perfect cube, the issue lies with the coefficient 215. Thus, to make the monomial a perfect cube, the number 215 needs to be changed.
1. Examine the exponents: For a monomial to be a perfect cube, all exponents in the expression, including the coefficient, must be multiples of 3.
- The exponent of [tex]\(x\)[/tex] is 18. Since 18 is a multiple of 3 (because [tex]\(18 \div 3 = 6\)[/tex]), [tex]\(x^{18}\)[/tex] is already a perfect cube.
- The exponent of [tex]\(y\)[/tex] is 3. Since 3 is a multiple of 3, [tex]\(y^3\)[/tex] is already a perfect cube.
- The exponent of [tex]\(z\)[/tex] is 21. Since 21 is a multiple of 3 (because [tex]\(21 \div 3 = 7\)[/tex]), [tex]\(z^{21}\)[/tex] is already a perfect cube.
2. Examine the coefficient (215): For the entire monomial to be a perfect cube, 215 must itself be a perfect cube. Let's consider its prime factorization:
- The prime factorization of 215 is [tex]\(5 \times 43\)[/tex].
A number is a perfect cube if each prime factor's exponent in the factorization is a multiple of 3. Here, [tex]\(5^1\)[/tex] and [tex]\(43^1\)[/tex] both have exponents of 1, which are not multiples of 3. Therefore, 215 is not a perfect cube.
Since the exponents of the variables [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] are already suitable for a perfect cube, the issue lies with the coefficient 215. Thus, to make the monomial a perfect cube, the number 215 needs to be changed.
