if y is a positive integer, for how many different values of y is $sqrt3{\frac{144}{y}}$ a whole…

Explanation:

Step1: Prime - factorize 144

$144 = 2^{4}\times3^{2}$

Step2: Let $\sqrt[3]{\frac{144}{y}}=n$, where $n$ is a whole - number. Then $\frac{144}{y}=n^{3}$.

So, $y=\frac{144}{n^{3}}$.

Step3: Analyze the cube - factors of 144

For $\frac{144}{y}$ to be a perfect cube, we need to make the exponents of prime factors of 144 divisible by 3. The prime - factorization of 144 is $2^{4}\times3^{2}$. We want to find positive - integer values of $y$ such that $\frac{144}{y}$ is a perfect cube. Let $y = 2^{a}\times3^{b}$. Then $\frac{144}{y}=\frac{2^{4}\times3^{2}}{2^{a}\times3^{b}}=2^{4 - a}\times3^{2 - b}$. For $2^{4 - a}\times3^{2 - b}$ to be a perfect cube, $4−a$ and $2 - b$ must be non - negative multiples of 3. For the exponent of 2: If $4−a = 0$ (a multiple of 3), then $a = 4$; if $4−a=3$, then $a = 1$. For the exponent of 3: If $2−b = 0$, then $b = 2$; if $2−b=- 1$ (not valid as $b$ is a non - negative integer). The possible values of $y$ are when we consider the combinations of factors to make the quotient a perfect cube. The factors of 144 are considered in terms of making the exponents of prime factors divisible by 3. The values of $y$ for which $\frac{144}{y}$ is a perfect cube are $y = 144$ (since $\sqrt[3]{\frac{144}{144}} = 1$) and $y = 18$ (because $\frac{144}{18}=8=2^{3}$).

Answer:

B. 2