College

Which number in the monomial [tex]$215 x^{18} y^3 z^{21}$[/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]\(215 x^{18} y^3 z^{21}\)[/tex] needs to be changed to make it a perfect cube, we need to check if all parts of the monomial are perfect cubes.

A perfect cube, by definition, means that the exponents (or powers) of each component in the expression should be multiples of 3. This includes the coefficient and the variables with their exponents.

Let's analyze each component:

1. Coefficient (215):
- We need to check if 215 is a perfect cube. This means looking into its prime factorization and ensuring each prime factor has an exponent that is a multiple of 3.
- The prime factorization of 215 is [tex]\(5 \times 43\)[/tex]. Neither 5 nor 43 are perfect cubes themselves and their exponents (1 each) are not multiples of 3.
- Hence, 215 is not a perfect cube.

2. Variable [tex]\(x^{18}\)[/tex]:
- The exponent of [tex]\(x\)[/tex] is 18. Since 18 is a multiple of 3 (18 divided by 3 equals 6 with no remainder), [tex]\(x^{18}\)[/tex] is a perfect cube.

3. Variable [tex]\(y^3\)[/tex]:
- The exponent of [tex]\(y\)[/tex] is 3. Since 3 is a multiple of 3, [tex]\(y^3\)[/tex] is a perfect cube.

4. Variable [tex]\(z^{21}\)[/tex]:
- The exponent of [tex]\(z\)[/tex] is 21. Since 21 is a multiple of 3 (21 divided by 3 equals 7 with no remainder), [tex]\(z^{21}\)[/tex] is a perfect cube.

From this analysis, the coefficients and exponents that are perfect cubes are:
- [tex]\((x^{18})\)[/tex] and [tex]\((y^3)\)[/tex] and [tex]\((z^{21})\)[/tex] are all perfect cubes because their exponents are multiples of 3.

Therefore, the part that needs to be adjusted is the coefficient 215. It is not a perfect cube because its prime factors do not have exponents that are multiples of 3. To make the entire monomial a perfect cube, the number 215 needs to be changed.