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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 ensure that the coefficient and all the exponents are divisible by 3.
Let's evaluate each component:
1. Examine the Coefficient:
- The coefficient is [tex]\(215\)[/tex]. For it to be a perfect cube, it has to be a number that equals [tex]\(k^3\)[/tex] for some integer [tex]\(k\)[/tex]. Upon examination, [tex]\(215\)[/tex] is not a perfect cube.
2. Examine the Exponents:
- For [tex]\(x^{18}\)[/tex]: The exponent is [tex]\(18\)[/tex]. Since 18 is divisible by 3, [tex]\(x^{18}\)[/tex] is already a perfect cube.
- For [tex]\(y^3\)[/tex]: The exponent is [tex]\(3\)[/tex]. Since 3 is divisible by 3, [tex]\(y^3\)[/tex] is also a perfect cube.
- For [tex]\(z^{21}\)[/tex]: The exponent is [tex]\(21\)[/tex]. Since 21 is divisible by 3, [tex]\(z^{21}\)[/tex] is similarly a perfect cube.
From the assessment above, all the exponents on the variables [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] are already perfect cubes because they are divisible by 3. However, the coefficient [tex]\(215\)[/tex] needs change to become a perfect cube.
Therefore, we need to change the coefficient, which is [tex]\(215\)[/tex]. To achieve this, it can be rounded to the nearest cube, which would be [tex]\(216\)[/tex], as [tex]\(216 = 6^3\)[/tex].
Thus, the number in the monomial [tex]\(215 x^{18} y^3 z^{21}\)[/tex] that needs to be changed to make the whole expression a perfect cube is 215.
Let's evaluate each component:
1. Examine the Coefficient:
- The coefficient is [tex]\(215\)[/tex]. For it to be a perfect cube, it has to be a number that equals [tex]\(k^3\)[/tex] for some integer [tex]\(k\)[/tex]. Upon examination, [tex]\(215\)[/tex] is not a perfect cube.
2. Examine the Exponents:
- For [tex]\(x^{18}\)[/tex]: The exponent is [tex]\(18\)[/tex]. Since 18 is divisible by 3, [tex]\(x^{18}\)[/tex] is already a perfect cube.
- For [tex]\(y^3\)[/tex]: The exponent is [tex]\(3\)[/tex]. Since 3 is divisible by 3, [tex]\(y^3\)[/tex] is also a perfect cube.
- For [tex]\(z^{21}\)[/tex]: The exponent is [tex]\(21\)[/tex]. Since 21 is divisible by 3, [tex]\(z^{21}\)[/tex] is similarly a perfect cube.
From the assessment above, all the exponents on the variables [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] are already perfect cubes because they are divisible by 3. However, the coefficient [tex]\(215\)[/tex] needs change to become a perfect cube.
Therefore, we need to change the coefficient, which is [tex]\(215\)[/tex]. To achieve this, it can be rounded to the nearest cube, which would be [tex]\(216\)[/tex], as [tex]\(216 = 6^3\)[/tex].
Thus, the number in the monomial [tex]\(215 x^{18} y^3 z^{21}\)[/tex] that needs to be changed to make the whole expression a perfect cube is 215.
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