Question

which expression is equivalent to $a^{1/3} + \frac{2}{a^{2/3}}$?\
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a. $\frac{a^{2/9}+2}{a^{2/3}}$\
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b. $\frac{2}{a^{1/3}}$\
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c. $\frac{a^{1/3}+2}{a^{2/3}}$\
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d. $\frac{a + 2}{a^{2/3}}$\
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e. $\frac{a^{2/3}+2}{a}$

Explanation:

Step1: Find a common denominator

To add \(a^{1/3}\) and \(\frac{2}{a^{2/3}}\), we need a common denominator. The common denominator here will be \(a^{2/3}\) because \(a^{1/3}\) can be rewritten with \(a^{2/3}\) as the denominator by multiplying the numerator and denominator by \(a^{1/3}\) (using the rule \(x^m \cdot x^n = x^{m + n}\)).

So, \(a^{1/3}=\frac{a^{1/3}\cdot a^{1/3}}{a^{1/3}}?\) Wait, no. Wait, to get a denominator of \(a^{2/3}\) for \(a^{1/3}\), we multiply numerator and denominator by \(a^{1/3}\):

\(a^{1/3}=\frac{a^{1/3}\cdot a^{1/3}}{a^{1/3}}\)? No, that's not right. Wait, \(a^{1/3}=\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}\) because when we multiply \(a^{1/3}\) and \(a^{2/3}\), we add the exponents: \(\frac{1}{3}+\frac{2}{3} = 1\), so \(a^{1/3}\cdot a^{2/3}=a^1 = a\). So:

\(a^{1/3}=\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}=\frac{a^{1}}{a^{2/3}}=\frac{a}{a^{2/3}}\)? Wait, no, that's not what we want. Wait, maybe I made a mistake. Let's start over.

The expression is \(a^{1/3}+\frac{2}{a^{2/3}}\). To combine these two terms, we need a common denominator. The second term has a denominator of \(a^{2/3}\), so we can rewrite the first term \(a^{1/3}\) with a denominator of \(a^{2/3}\) by multiplying the numerator and denominator by \(a^{1/3}\) (because \(a^{1/3}\cdot a^{1/3}=a^{2/3}\)? Wait, no: \(a^{1/3}\cdot a^{1/3}=a^{\frac{1}{3}+\frac{1}{3}}=a^{2/3}\). Wait, no, \(\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\), so \(a^{1/3}\cdot a^{1/3}=a^{2/3}\). So:

\(a^{1/3}=\frac{a^{1/3}\cdot a^{1/3}}{a^{1/3}}\)? No, that's not. Wait, to get a denominator of \(a^{2/3}\) for \(a^{1/3}\), we multiply numerator and denominator by \(a^{1/3}\):

\(a^{1/3}=\frac{a^{1/3}\cdot a^{1/3}}{a^{1/3}}\)? No, that's not. Wait, \(a^{1/3}=\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}\) because \(a^{1/3}\cdot a^{2/3}=a^{\frac{1}{3}+\frac{2}{3}}=a^1 = a\). So:

\(a^{1/3}=\frac{a}{a^{2/3}}\)

Then, the original expression \(a^{1/3}+\frac{2}{a^{2/3}}\) becomes \(\frac{a}{a^{2/3}}+\frac{2}{a^{2/3}}\)

Wait, but that would be \(\frac{a + 2}{a^{2/3}}\), but that's option D. But the selected option is C. Wait, maybe I made a mistake.

Wait, no, maybe the initial approach is wrong. Let's check the options.

Wait, the expression is \(a^{1/3}+\frac{2}{a^{2/3}}\). Let's rewrite \(a^{1/3}\) as \(\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}\)? No, that would be \(\frac{a^{1/3 + 2/3}}{a^{2/3}}=\frac{a^1}{a^{2/3}}=\frac{a}{a^{2/3}}\). Then adding \(\frac{2}{a^{2/3}}\) gives \(\frac{a + 2}{a^{2/3}}\), which is option D. But the selected option is C. Wait, maybe the problem is written differently. Wait, the original problem is \(a^{1/3}+\frac{2}{a^{2/3}}\). Let's check the options again.

Option C is \(\frac{a^{1/3}+2}{a^{2/3}}\). Let's see if that's equivalent. Let's combine the terms:

\(a^{1/3}+\frac{2}{a^{2/3}}=\frac{a^{1/3}\cdot a^{2/3}+2}{a^{2/3}}\)? No, that's not. Wait, no, to combine \(a^{1/3}\) and \(\frac{2}{a^{2/3}}\), we need a common denominator. The common denominator is \(a^{2/3}\). So, \(a^{1/3}=\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}\)? No, that's multiplying by \(a^{2/3}\) in the numerator and denominator. Wait, no, \(a^{1/3}=\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}\) would be \(\frac{a^{1/3 + 2/3}}{a^{2/3}}=\frac{a^1}{a^{2/3}}=\frac{a}{a^{2/3}}\). Then adding \(\frac{2}{a^{2/3}}\) gives \(\frac{a + 2}{a^{2/3}}\) (option D). But the selected option is C. Wait, maybe the problem is \(a^{1/3}+\frac{2}{a^{2/3}}\) and the options are as given. Wait, maybe I made a mistake in the exponent rules.

Wait, let's check option C: \(\frac{a^{1/3}+2}{a^{2/3}}\). Let's expand this: \(\frac{a^{1/3}}{a^{2/3}}+\frac{2}{a^{2/3}}=a^{1/3 - 2/3}+\frac{2}{a^{2/3}}=a^{-1/3}+\frac{2}{a^{2/3}}\), which is not the original expression. Wait, that can't be.

Wait, maybe the original problem is \(a^{1/3}+\frac{2}{a^{2/3}}\) and the correct answer is D? But the user has selected C. Wait, maybe there's a typo. Wait, let's re-express the original expression:

\(a^{1/3}+\frac{2}{a^{2/3}}\). Let's get a common denominator of \(a^{2/3}\):

\(a^{1/3}=\frac{a^{1/3}\cdot a^{1/3}}{a^{1/3}}\)? No, that's not. Wait, no, \(a^{1/3}=\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}\) is incorrect. Wait, the correct way to get a common denominator is to note that \(a^{1/3}=\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}\) is wrong. Wait, the exponent rule for division is \(x^m / x^n = x^{m - n}\). So, \(a^{1/3}=a^{1/3}/1\), and to get a denominator of \(a^{2/3}\), we multiply numerator and denominator by \(a^{2/3}\):

\(a^{1/3}=\frac{a^{1/3}\cdot a^{2/3}}{a^{2/3}}=\frac{a^{1/3 + 2/3}}{a^{2/3}}=\frac{a^1}{a^{2/3}}=\frac{a}{a^{2/3}}\)

Then, adding \(\frac{2}{a^{2/3}}\) gives \(\frac{a + 2}{a^{2/3}}\), which is option D. But the selected option is C. Wait, maybe the problem was supposed to be \(a^{1/3}+\frac{2}{a^{1/3}}\) instead of \(a^{2/3}\)? Let's check. If the denominator was \(a^{1/3}\), then:

\(a^{1/3}+\frac{2}{a^{1/3}}=\frac{a^{1/3}\cdot a^{1/3}+2}{a^{1/3}}=\frac{a^{2/3}+2}{a^{1/3}}\), which is not an option.

Alternatively, maybe the original expression is \(a^{1/3}+\frac{2}{a^{2/3}}\) and the correct answer is D, but the user selected C. Wait, maybe I made a mistake. Let's check the options again:

A. \(\frac{a^{2/9}+2}{a^{2/3}}\) – no, that doesn't make sense.

B. \(\frac{2}{a^{1/3}}\) – no, that's not equivalent.

C. \(\frac{a^{1/3}+2}{a^{2/3}}\) – let's expand this: \(\frac{a^{1/3}}{a^{2/3}}+\frac{2}{a^{2/3}}=a^{1/3 - 2/3}+\frac{2}{a^{2/3}}=a^{-1/3}+\frac{2}{a^{2/3}}\), which is not the original expression.

D. \(\frac{a + 2}{a^{2/3}}\) – expanding this: \(\frac{a}{a^{2/3}}+\frac{2}{a^{2/3}}=a^{1 - 2/3}+\frac{2}{a^{2/3}}=a^{1/3}+\frac{2}{a^{2/3}}\), which matches the original expression.

E. \(\frac{a^{2/3}+2}{a}\) – no.

So the correct answer should be D, but the user has selected C. Maybe there's a mistake in the problem or the selection. But according to the calculation, the equivalent expression is \(\frac{a + 2}{a^{2/3}}\), which is option D.

Wait, but let's re-express the original expression:

Original: \(a^{1/3}+\frac{2}{a^{2/3}}\)

Let's compute option D: \(\frac{a + 2}{a^{2/3}}=\frac{a}{a^{2/3}}+\frac{2}{a^{2/3}}=a^{1 - 2/3}+\frac{2}{a^{2/3}}=a^{1/3}+\frac{2}{a^{2/3}}\), which is the original expression. So option D is correct.

But the user has selected C. Maybe there's a typo in the problem. Alternatively, maybe the original expression is \(a^{1/3}+\frac{2}{a^{1/3}}\), but that's not what's written.

Alternatively, maybe I made a mistake in the exponent rules. Let's verify:

\(a^{1/3}\) is the cube root of \(a\), and \(a^{2/3}\) is the cube root of \(a^2\). So, \(a^{1/3}+\frac{2}{a^{2/3}}\). Let's take \(a = 8\) (so \(a^{1/3}=2\), \(a^{2/3}=4\)):

Original expression: \(2 + \frac{2}{4}=2 + 0.5 = 2.5\)

Option C: \(\frac{2 + 2}{4}=\frac{4}{4}=1\) – not equal.

Option D: \(\frac{8 + 2}{4}=\frac{10}{4}=2.5\) – equal.

Ah! So when \(a = 8\), original expression is 2.5, option D is 2.5, option C is 1. So the correct answer is D. So the user's selection of C is incorrect. The correct answer is D.

Answer:

D. \(\frac{a + 2}{a^{2/3}}\)