Answer :
A monomial is a perfect cube if every exponent—including the exponents of the variables and of the prime factors of the numerical coefficient—is a multiple of 3.
1. For the monomial
$$
215 x^{18} y^3 z^{21},
$$
we first check the exponents of the variables:
- The exponent of $x$ is 18, and $18 = 3 \times 6$.
- The exponent of $y$ is 3, and $3 = 3 \times 1$.
- The exponent of $z$ is 21, and $21 = 3 \times 7$.
Since $18$, $3$, and $21$ are all divisible by 3, the variable parts already form perfect cubes.
2. Next, consider the numerical coefficient $215$. For $215$ to be a perfect cube, when we express $215$ as a product of its prime factors, each prime’s exponent must be a multiple of 3.
Factorize $215$:
$$
215 = 5^1 \times 43^1.
$$
Here, the exponents of $5$ and $43$ are 1, which are not multiples of 3.
Since the variable exponents are appropriate, the number that must be changed to achieve a perfect cube is $215$.
1. For the monomial
$$
215 x^{18} y^3 z^{21},
$$
we first check the exponents of the variables:
- The exponent of $x$ is 18, and $18 = 3 \times 6$.
- The exponent of $y$ is 3, and $3 = 3 \times 1$.
- The exponent of $z$ is 21, and $21 = 3 \times 7$.
Since $18$, $3$, and $21$ are all divisible by 3, the variable parts already form perfect cubes.
2. Next, consider the numerical coefficient $215$. For $215$ to be a perfect cube, when we express $215$ as a product of its prime factors, each prime’s exponent must be a multiple of 3.
Factorize $215$:
$$
215 = 5^1 \times 43^1.
$$
Here, the exponents of $5$ and $43$ are 1, which are not multiples of 3.
Since the variable exponents are appropriate, the number that must be changed to achieve a perfect cube is $215$.
