Answer :
A monomial is a perfect cube if every exponent in the expression is a multiple of 3, and the numerical coefficient is also a perfect cube.
First, consider the variable factors in the monomial
[tex]$$215 \, x^{18} y^3 z^{21}.$$[/tex]
- The exponent on [tex]\(x\)[/tex] is [tex]\(18\)[/tex], and since [tex]\(18 = 3 \times 6\)[/tex], it is a multiple of 3.
- The exponent on [tex]\(y\)[/tex] is [tex]\(3\)[/tex], and [tex]\(3 = 3 \times 1\)[/tex], so it is a multiple of 3.
- The exponent on [tex]\(z\)[/tex] is [tex]\(21\)[/tex], and [tex]\(21 = 3 \times 7\)[/tex], which is also a multiple of 3.
Next, examine the numerical coefficient [tex]\(215\)[/tex]. A number is a perfect cube if there exists an integer [tex]\(n\)[/tex] such that
[tex]$$n^3 = \text{coefficient}.$$[/tex]
Looking at nearby cubes:
[tex]$$5^3 = 125 \quad \text{and} \quad 6^3 = 216.$$[/tex]
Since [tex]\(215\)[/tex] is not equal to [tex]\(216\)[/tex] and lies between [tex]\(5^3\)[/tex] and [tex]\(6^3\)[/tex], it is not a perfect cube.
Thus, to transform
[tex]$$215 \, x^{18} y^3 z^{21}$$[/tex]
into a perfect cube, the coefficient [tex]\(215\)[/tex] must be changed.
The answer is [tex]\(\boxed{215}\)[/tex].
First, consider the variable factors in the monomial
[tex]$$215 \, x^{18} y^3 z^{21}.$$[/tex]
- The exponent on [tex]\(x\)[/tex] is [tex]\(18\)[/tex], and since [tex]\(18 = 3 \times 6\)[/tex], it is a multiple of 3.
- The exponent on [tex]\(y\)[/tex] is [tex]\(3\)[/tex], and [tex]\(3 = 3 \times 1\)[/tex], so it is a multiple of 3.
- The exponent on [tex]\(z\)[/tex] is [tex]\(21\)[/tex], and [tex]\(21 = 3 \times 7\)[/tex], which is also a multiple of 3.
Next, examine the numerical coefficient [tex]\(215\)[/tex]. A number is a perfect cube if there exists an integer [tex]\(n\)[/tex] such that
[tex]$$n^3 = \text{coefficient}.$$[/tex]
Looking at nearby cubes:
[tex]$$5^3 = 125 \quad \text{and} \quad 6^3 = 216.$$[/tex]
Since [tex]\(215\)[/tex] is not equal to [tex]\(216\)[/tex] and lies between [tex]\(5^3\)[/tex] and [tex]\(6^3\)[/tex], it is not a perfect cube.
Thus, to transform
[tex]$$215 \, x^{18} y^3 z^{21}$$[/tex]
into a perfect cube, the coefficient [tex]\(215\)[/tex] must be changed.
The answer is [tex]\(\boxed{215}\)[/tex].
