To make the monomial [tex]\(125 x^{18} y^3 z^{25}\)[/tex] a perfect cube, we need to ensure that all components are perfect cubes.
1. Considering the coefficient [tex]\(125\)[/tex]:
- The number [tex]\(125\)[/tex] is already a perfect cube since [tex]\(125 = 5^3\)[/tex].
2. Considering the variable [tex]\(x^{18}\)[/tex]:
- The exponent of [tex]\(x\)[/tex] is [tex]\(18\)[/tex]. To be a perfect cube, the exponent must be divisible by [tex]\(3\)[/tex]. Since [tex]\(18\)[/tex] is divisible by [tex]\(3\)[/tex], [tex]\(x^{18}\)[/tex] is a perfect cube.
3. Considering the variable [tex]\(y^3\)[/tex]:
- The exponent of [tex]\(y\)[/tex] is [tex]\(3\)[/tex]. To be a perfect cube, the exponent must be divisible by [tex]\(3\)[/tex]. Since [tex]\(3\)[/tex] is divisible by [tex]\(3\)[/tex], [tex]\(y^3\)[/tex] is already a perfect cube.
4. Considering the variable [tex]\(z^{25}\)[/tex]:
- The exponent of [tex]\(z\)[/tex] is [tex]\(25\)[/tex]. To be a perfect cube, the exponent must be divisible by [tex]\(3\)[/tex]. Since [tex]\(25\)[/tex] is not divisible by [tex]\(3\)[/tex], [tex]\(z^{25}\)[/tex] is not a perfect cube.
To make [tex]\(z^{25}\)[/tex] a perfect cube, we must change the exponent to the nearest lower multiple of [tex]\(3\)[/tex], which is [tex]\(24\)[/tex]. Thus, [tex]\(z^{25}\)[/tex] should be changed to [tex]\(z^{24}\)[/tex].
Therefore, the number that needs to be changed is:
[tex]\[ 25 \][/tex]
