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, let's follow these steps:
1. Understanding Perfect Cubes:
- For a monomial to be a perfect cube, all the exponents in its prime factorization, including the constant term, must be multiples of 3.
2. Identify Exponents:
- The exponent for [tex]\(x\)[/tex] is 18, which is already a multiple of 3.
- The exponent for [tex]\(y\)[/tex] is 3, which is also a multiple of 3.
- The exponent for [tex]\(z\)[/tex] is 21, which is already a multiple of 3.
3. Consider the Coefficient:
- The number 215 (the coefficient) also needs to be a perfect cube. To check if 215 is a perfect cube, factor it into prime factors:
- The prime factorization of 215 is [tex]\(5 \times 43\)[/tex]. Neither [tex]\(5\)[/tex] nor [tex]\(43\)[/tex] is a cube, nor can they be made into a cube by multiplying 215 by any factor except its square.
4. Making 215 a Perfect Cube:
- To change 215 into a perfect cube, each prime number in its factorization would need to appear in triples. Therefore, since 215 = [tex]\(5 \times 43\)[/tex], neither 5 nor 43 can form cubes alone unless squared.
- Multiplying 215 by itself, (215 \times 215=215^2), gives [tex]\(5^3 \times 43^3\)[/tex] after simplification, making it a perfect cube.
Therefore, the number that needs to be changed in the monomial to make the entire monomial a perfect cube is 215.
1. Understanding Perfect Cubes:
- For a monomial to be a perfect cube, all the exponents in its prime factorization, including the constant term, must be multiples of 3.
2. Identify Exponents:
- The exponent for [tex]\(x\)[/tex] is 18, which is already a multiple of 3.
- The exponent for [tex]\(y\)[/tex] is 3, which is also a multiple of 3.
- The exponent for [tex]\(z\)[/tex] is 21, which is already a multiple of 3.
3. Consider the Coefficient:
- The number 215 (the coefficient) also needs to be a perfect cube. To check if 215 is a perfect cube, factor it into prime factors:
- The prime factorization of 215 is [tex]\(5 \times 43\)[/tex]. Neither [tex]\(5\)[/tex] nor [tex]\(43\)[/tex] is a cube, nor can they be made into a cube by multiplying 215 by any factor except its square.
4. Making 215 a Perfect Cube:
- To change 215 into a perfect cube, each prime number in its factorization would need to appear in triples. Therefore, since 215 = [tex]\(5 \times 43\)[/tex], neither 5 nor 43 can form cubes alone unless squared.
- Multiplying 215 by itself, (215 \times 215=215^2), gives [tex]\(5^3 \times 43^3\)[/tex] after simplification, making it a perfect cube.
Therefore, the number that needs to be changed in the monomial to make the entire monomial a perfect cube is 215.
