College

Which number in the monomial [tex]215 x^{18} y^3 \xi^{21}[/tex] needs to be changed to make it a perfect cube?

A. 3
B. 18
C. 21
D. 215

Answer :

To determine which number in the monomial [tex]\(215 x^{18} y^3 \xi^{21}\)[/tex] needs to be changed to make it a perfect cube, let's go through the components one by one:

1. Exponents of the variables:
- For the monomial to be a perfect cube, all exponents should be multiples of 3.
- [tex]\(x^{18}\)[/tex]: The exponent is 18, which is a multiple of 3, so it’s fine.
- [tex]\(y^3\)[/tex]: The exponent is 3, which is also a multiple of 3, so it works.
- [tex]\(\xi^{21}\)[/tex]: The exponent is 21, which is a multiple of 3 as well, so it is okay.

2. The coefficient (number):
- We need the coefficient, 215, to be a perfect cube.
- A perfect cube is a number that can be expressed as another integer raised to the power of 3.

To check if 215 is a perfect cube, we factor it:
- Prime factorization of 215: It is [tex]\(5 \times 43\)[/tex].
- Both 5 and 43 have an exponent of 1, none of which are multiples of 3.

Since the exponents in the factorization of 215 don’t form a perfect cube (they’re not multiples of 3), the coefficient 215 must be changed to make the monomial a perfect cube.

Therefore, the number in the monomial that needs to be changed to make it a perfect cube is 215.