Answer :
To determine which number in the monomial [tex]\(215 x^{18} y^3 z^{21}\)[/tex] should be changed to make it a perfect cube, let's look at the requirements for a monomial to be a perfect cube:
1. Understanding Perfect Cubes: A monomial is a perfect cube if each of its components (the coefficient and the variable powers) is a perfect cube.
2. Analyze the Numbers:
- For the Coefficient (215): The prime factorization of 215 is [tex]\(5 \times 43\)[/tex]. Neither 5 nor 43 is a perfect cube, and 215 cannot be changed simply by exponentiation to make it a perfect cube because it would require complete re-factorization.
3. Analyze the Exponents:
- For [tex]\(x^{18}\)[/tex]: The exponent 18 is already a multiple of 3 (since [tex]\(18 \div 3 = 6\)[/tex]). Therefore, [tex]\(x^{18}\)[/tex] is a perfect cube.
- For [tex]\(y^3\)[/tex]: The exponent 3 is a multiple of 3. Thus, [tex]\(y^3\)[/tex] is also a perfect cube.
- For [tex]\(z^{21}\)[/tex]: The exponent 21 is a multiple of 3 (since [tex]\(21 \div 3 = 7\)[/tex]). So, [tex]\(z^{21}\)[/tex] is a perfect cube.
4. Conclusion: The component that prevents the monomial from being a perfect cube is the coefficient 215. Therefore, 215 is the number that needs to be changed to make the monomial a perfect cube.
1. Understanding Perfect Cubes: A monomial is a perfect cube if each of its components (the coefficient and the variable powers) is a perfect cube.
2. Analyze the Numbers:
- For the Coefficient (215): The prime factorization of 215 is [tex]\(5 \times 43\)[/tex]. Neither 5 nor 43 is a perfect cube, and 215 cannot be changed simply by exponentiation to make it a perfect cube because it would require complete re-factorization.
3. Analyze the Exponents:
- For [tex]\(x^{18}\)[/tex]: The exponent 18 is already a multiple of 3 (since [tex]\(18 \div 3 = 6\)[/tex]). Therefore, [tex]\(x^{18}\)[/tex] is a perfect cube.
- For [tex]\(y^3\)[/tex]: The exponent 3 is a multiple of 3. Thus, [tex]\(y^3\)[/tex] is also a perfect cube.
- For [tex]\(z^{21}\)[/tex]: The exponent 21 is a multiple of 3 (since [tex]\(21 \div 3 = 7\)[/tex]). So, [tex]\(z^{21}\)[/tex] is a perfect cube.
4. Conclusion: The component that prevents the monomial from being a perfect cube is the coefficient 215. Therefore, 215 is the number that needs to be changed to make the monomial a perfect cube.
