Answer :
To determine which number must change, we first recall that in a perfect cube every exponent in the monomial, including the exponents in the prime factorization of its numerical coefficient, must be a multiple of 3.
The given monomial is
[tex]$$
215\, x^{18}\, y^3\, z^{21}.
$$[/tex]
Step 1. Check the variable exponents:
- For [tex]$x^{18}$[/tex], the exponent is [tex]$18$[/tex], and since [tex]$18 \div 3 = 6$[/tex], it is a multiple of 3.
- For [tex]$y^3$[/tex], the exponent is [tex]$3$[/tex], and [tex]$3 \div 3 = 1$[/tex], so it is a multiple of 3.
- For [tex]$z^{21}$[/tex], the exponent is [tex]$21$[/tex], and [tex]$21 \div 3 = 7$[/tex], so it is also a multiple of 3.
Step 2. Check the coefficient:
The coefficient is [tex]$215$[/tex]. We first factor it into primes:
[tex]$$
215 = 5 \times 43.
$$[/tex]
Here, the exponents for both prime factors [tex]$5$[/tex] and [tex]$43$[/tex] are [tex]$1$[/tex]. However, [tex]$1$[/tex] is not a multiple of [tex]$3$[/tex]. For the monomial to be a perfect cube, each prime factor in the coefficient must appear with an exponent that is a multiple of 3 (for example, [tex]$5^3$[/tex], [tex]$43^3$[/tex], etc.).
Since the exponents exist as [tex]$1$[/tex] (and [tex]$1$[/tex] is not divisible by [tex]$3$[/tex]), the constant [tex]$215$[/tex] does not meet the perfect cube requirement.
Conclusion:
Among the numbers provided ([tex]$18$[/tex], [tex]$215$[/tex], [tex]$21$[/tex], [tex]$3$[/tex]), it is only the coefficient [tex]$215$[/tex] that needs to be changed in order to make the monomial a perfect cube.
Thus, the number to be changed is [tex]$\boxed{215}$[/tex].
The given monomial is
[tex]$$
215\, x^{18}\, y^3\, z^{21}.
$$[/tex]
Step 1. Check the variable exponents:
- For [tex]$x^{18}$[/tex], the exponent is [tex]$18$[/tex], and since [tex]$18 \div 3 = 6$[/tex], it is a multiple of 3.
- For [tex]$y^3$[/tex], the exponent is [tex]$3$[/tex], and [tex]$3 \div 3 = 1$[/tex], so it is a multiple of 3.
- For [tex]$z^{21}$[/tex], the exponent is [tex]$21$[/tex], and [tex]$21 \div 3 = 7$[/tex], so it is also a multiple of 3.
Step 2. Check the coefficient:
The coefficient is [tex]$215$[/tex]. We first factor it into primes:
[tex]$$
215 = 5 \times 43.
$$[/tex]
Here, the exponents for both prime factors [tex]$5$[/tex] and [tex]$43$[/tex] are [tex]$1$[/tex]. However, [tex]$1$[/tex] is not a multiple of [tex]$3$[/tex]. For the monomial to be a perfect cube, each prime factor in the coefficient must appear with an exponent that is a multiple of 3 (for example, [tex]$5^3$[/tex], [tex]$43^3$[/tex], etc.).
Since the exponents exist as [tex]$1$[/tex] (and [tex]$1$[/tex] is not divisible by [tex]$3$[/tex]), the constant [tex]$215$[/tex] does not meet the perfect cube requirement.
Conclusion:
Among the numbers provided ([tex]$18$[/tex], [tex]$215$[/tex], [tex]$21$[/tex], [tex]$3$[/tex]), it is only the coefficient [tex]$215$[/tex] that needs to be changed in order to make the monomial a perfect cube.
Thus, the number to be changed is [tex]$\boxed{215}$[/tex].
