Answer :
To determine when a monomial is a perfect cube, every factor in the monomial must have its exponent be a multiple of 3. In the monomial
[tex]$$
215 x^{18} y^3 z^{21},
$$[/tex]
we need to ensure that:
1. The exponents on the variables [tex]$x$[/tex], [tex]$y$[/tex], and [tex]$z$[/tex] are multiples of 3.
2. The coefficient (a constant number) is also a perfect cube; that is, in its prime factorization, each prime’s exponent must be a multiple of 3.
Step 1. Check the Variable Exponents
- For [tex]$x^{18}$[/tex], the exponent is [tex]$18$[/tex]. Since
[tex]$$
18 \div 3 = 6,
$$[/tex]
and [tex]$6$[/tex] is an integer, [tex]$18$[/tex] is a multiple of 3.
- For [tex]$y^{3}$[/tex], the exponent is [tex]$3$[/tex]. Since
[tex]$$
3 \div 3 = 1,
$$[/tex]
[tex]$3$[/tex] is a multiple of 3.
- For [tex]$z^{21}$[/tex], the exponent is [tex]$21$[/tex]. Since
[tex]$$
21 \div 3 = 7,
$$[/tex]
[tex]$21$[/tex] is also a multiple of 3.
Thus, all the variable exponents satisfy the condition needed for a perfect cube.
Step 2. Check the Coefficient
The coefficient is [tex]$215$[/tex]. A number is a perfect cube if, when expressed as a product of primes, all the exponents in the factorization are multiples of 3. The prime factorization of [tex]$215$[/tex] is
[tex]$$
215 = 5 \times 43.
$$[/tex]
Here, both [tex]$5$[/tex] and [tex]$43$[/tex] appear with an exponent of [tex]$1$[/tex], and since [tex]$1$[/tex] is not a multiple of [tex]$3$[/tex], the number [tex]$215$[/tex] is not a perfect cube.
Conclusion
The variable parts are already perfect cubes because [tex]$18$[/tex], [tex]$3$[/tex], and [tex]$21$[/tex] are all divisible by [tex]$3$[/tex]. However, the coefficient [tex]$215$[/tex] is not a perfect cube. Therefore, to make the entire monomial a perfect cube, the number that must be changed is the coefficient:
[tex]$$
\boxed{215}.
$$[/tex]
[tex]$$
215 x^{18} y^3 z^{21},
$$[/tex]
we need to ensure that:
1. The exponents on the variables [tex]$x$[/tex], [tex]$y$[/tex], and [tex]$z$[/tex] are multiples of 3.
2. The coefficient (a constant number) is also a perfect cube; that is, in its prime factorization, each prime’s exponent must be a multiple of 3.
Step 1. Check the Variable Exponents
- For [tex]$x^{18}$[/tex], the exponent is [tex]$18$[/tex]. Since
[tex]$$
18 \div 3 = 6,
$$[/tex]
and [tex]$6$[/tex] is an integer, [tex]$18$[/tex] is a multiple of 3.
- For [tex]$y^{3}$[/tex], the exponent is [tex]$3$[/tex]. Since
[tex]$$
3 \div 3 = 1,
$$[/tex]
[tex]$3$[/tex] is a multiple of 3.
- For [tex]$z^{21}$[/tex], the exponent is [tex]$21$[/tex]. Since
[tex]$$
21 \div 3 = 7,
$$[/tex]
[tex]$21$[/tex] is also a multiple of 3.
Thus, all the variable exponents satisfy the condition needed for a perfect cube.
Step 2. Check the Coefficient
The coefficient is [tex]$215$[/tex]. A number is a perfect cube if, when expressed as a product of primes, all the exponents in the factorization are multiples of 3. The prime factorization of [tex]$215$[/tex] is
[tex]$$
215 = 5 \times 43.
$$[/tex]
Here, both [tex]$5$[/tex] and [tex]$43$[/tex] appear with an exponent of [tex]$1$[/tex], and since [tex]$1$[/tex] is not a multiple of [tex]$3$[/tex], the number [tex]$215$[/tex] is not a perfect cube.
Conclusion
The variable parts are already perfect cubes because [tex]$18$[/tex], [tex]$3$[/tex], and [tex]$21$[/tex] are all divisible by [tex]$3$[/tex]. However, the coefficient [tex]$215$[/tex] is not a perfect cube. Therefore, to make the entire monomial a perfect cube, the number that must be changed is the coefficient:
[tex]$$
\boxed{215}.
$$[/tex]
