Question

which expression is equivalent to \\(\frac{a^{2/3}}{b^{1/3}} + \frac{b^{2/3}}{a^{1/3}}\\)?

a. \\(\frac{a + b}{a^{1/3}b^{1/3}}\\)

b. \\(\frac{a^{2/3} + b^{2/3}}{a^{1/3} + b^{1/3}}\\)

c. \\(\frac{a + b}{a^{1/3} + b^{1/3}}\\)

d. \\(\frac{a^{2/9} + b^{2/9}}{a^{1/3}b^{1/3}}\\)

e. \\(\frac{a^{2/3}b^{1/3} + a^{1/3}b^{2/3}}{a^{1/3}b^{1/3}}\\)

Explanation:

Step1: Find a common denominator

To add the two fractions \(\frac{a^{2/3}}{b^{1/3}}\) and \(\frac{b^{2/3}}{a^{1/3}}\), we need a common denominator. The common denominator of \(b^{1/3}\) and \(a^{1/3}\) is \(a^{1/3}b^{1/3}\).

Step2: Rewrite each fraction with the common denominator

For the first fraction \(\frac{a^{2/3}}{b^{1/3}}\), multiply the numerator and denominator by \(a^{1/3}\):
\[ \frac{a^{2/3} \cdot a^{1/3}}{b^{1/3} \cdot a^{1/3}}=\frac{a^{2/3 + 1/3}}{a^{1/3}b^{1/3}}=\frac{a^{1}}{a^{1/3}b^{1/3}}=\frac{a}{a^{1/3}b^{1/3}} \]
For the second fraction \(\frac{b^{2/3}}{a^{1/3}}\), multiply the numerator and denominator by \(b^{1/3}\):
\[ \frac{b^{2/3} \cdot b^{1/3}}{a^{1/3} \cdot b^{1/3}}=\frac{b^{2/3+1/3}}{a^{1/3}b^{1/3}}=\frac{b^{1}}{a^{1/3}b^{1/3}}=\frac{b}{a^{1/3}b^{1/3}} \]

Step3: Add the two fractions

Now that both fractions have the same denominator, we can add their numerators:
\[ \frac{a}{a^{1/3}b^{1/3}}+\frac{b}{a^{1/3}b^{1/3}}=\frac{a + b}{a^{1/3}b^{1/3}} \]
Wait, no, wait. Wait, actually, when we rewrite the first fraction: \(\frac{a^{2/3}}{b^{1/3}}=\frac{a^{2/3}\cdot a^{1/3}}{b^{1/3}\cdot a^{1/3}}=\frac{a^{2/3 + 1/3}}{a^{1/3}b^{1/3}}=\frac{a^{1}}{a^{1/3}b^{1/3}}\)? Wait, no, \(2/3+1/3 = 1\), so \(a^{2/3}\cdot a^{1/3}=a^{1}\), correct. And the second fraction: \(\frac{b^{2/3}}{a^{1/3}}=\frac{b^{2/3}\cdot b^{1/3}}{a^{1/3}\cdot b^{1/3}}=\frac{b^{2/3 + 1/3}}{a^{1/3}b^{1/3}}=\frac{b^{1}}{a^{1/3}b^{1/3}}\). Then adding them: \(\frac{a + b}{a^{1/3}b^{1/3}}\)? But wait, let's check option A: \(\frac{a + b}{a^{1/3}b^{1/3}}\), but wait, let's re - do the addition.

Wait, actually, the original expression is \(\frac{a^{2/3}}{b^{1/3}}+\frac{b^{2/3}}{a^{1/3}}\). Let's get a common denominator of \(a^{1/3}b^{1/3}\). So:

\[ \frac{a^{2/3}}{b^{1/3}}\times\frac{a^{1/3}}{a^{1/3}}+\frac{b^{2/3}}{a^{1/3}}\times\frac{b^{1/3}}{b^{1/3}}=\frac{a^{2/3}a^{1/3}+b^{2/3}b^{1/3}}{a^{1/3}b^{1/3}} \]

Using the exponent rule \(x^m\times x^n=x^{m + n}\), we have \(a^{2/3}a^{1/3}=a^{2/3+1/3}=a^{1}=a\) and \(b^{2/3}b^{1/3}=b^{2/3 + 1/3}=b^{1}=b\). So the numerator is \(a + b\) and the denominator is \(a^{1/3}b^{1/3}\). But wait, option A is \(\frac{a + b}{a^{1/3}b^{1/3}}\), but let's check the options again. Wait, option E is \(\frac{a^{2/3}b^{1/3}+a^{1/3}b^{2/3}}{a^{1/3}b^{1/3}}\). Wait, I made a mistake in the previous step. Let's correct it.

When we rewrite \(\frac{a^{2/3}}{b^{1/3}}\) with denominator \(a^{1/3}b^{1/3}\), we multiply numerator and denominator by \(a^{1/3}\), so we get \(\frac{a^{2/3}\times a^{1/3}}{b^{1/3}\times a^{1/3}}=\frac{a^{2/3 + 1/3}}{a^{1/3}b^{1/3}}=\frac{a^{1}}{a^{1/3}b^{1/3}}\)? No, that's wrong. Wait, \(a^{2/3}\times a^{1/3}=a^{(2 + 1)/3}=a^{1}\), but the numerator is \(a^{2/3}\times a^{1/3}=a^{1}\), but the correct numerator when we don't simplify the exponent first is \(a^{2/3}a^{1/3}\), and the second fraction \(\frac{b^{2/3}}{a^{1/3}}\) multiplied by \(\frac{b^{1/3}}{b^{1/3}}\) gives \(\frac{b^{2/3}\times b^{1/3}}{a^{1/3}\times b^{1/3}}=\frac{b^{2/3+1/3}}{a^{1/3}b^{1/3}}=\frac{b^{1}}{a^{1/3}b^{1/3}}\). But if we don't simplify the exponents in the numerator, the numerator of the sum is \(a^{2/3}a^{1/3}+b^{2/3}b^{1/3}=a^{2/3 + 1/3}+b^{2/3+1/3}=a + b\)? Wait, no, \(a^{2/3}a^{1/3}=a^{(2 + 1)/3}=a^{1}\), and \(b^{2/3}b^{1/3}=b^{1}\), so the numerator is \(a + b\) and the denominator is \(a^{1/3}b^{1/3}\). But option E's numerator is \(a^{2/3}b^{1/3}+a^{1/3}b^{2/3}\). Wait, let's factor the numerator of option E: \(a^{2/3}b^{1/3}+a^{1/3}b^{2/3}=a^{1/3}b^{1/3}(a^{1/3}+b^{1/3})\). And the denominator is \(a^{1/3}b^{1/3}\), so \(\frac{a^{1/3}b^{1/3}(a^{1/3}+b^{1/3})}{a^{1/3}b^{1/3}}=a^{1/3}+b^{1/3}\), which is not correct. Wait, I think I messed up the initial step.

Let's start over. The expression is \(\frac{a^{2/3}}{b^{1/3}}+\frac{b^{2/3}}{a^{1/3}}\). To add these two fractions, we find a common denominator, which is \(a^{1/3}b^{1/3}\). So:

\[ \frac{a^{2/3}}{b^{1/3}}+\frac{b^{2/3}}{a^{1/3}}=\frac{a^{2/3}\times a^{1/3}}{b^{1/3}\times a^{1/3}}+\frac{b^{2/3}\times b^{1/3}}{a^{1/3}\times b^{1/3}} \]

Simplify the numerators:

For the first term: \(a^{2/3}\times a^{1/3}=a^{(2/3)+(1/3)}=a^{1}=a\)

For the second term: \(b^{2/3}\times b^{1/3}=b^{(2/3)+(1/3)}=b^{1}=b\)

So the sum is \(\frac{a + b}{a^{1/3}b^{1/3}}\), which is option A? But wait, option E is \(\frac{a^{2/3}b^{1/3}+a^{1/3}b^{2/3}}{a^{1/3}b^{1/3}}\). Let's simplify option E's numerator: \(a^{2/3}b^{1/3}+a^{1/3}b^{2/3}=a^{1/3}b^{1/3}(a^{1/3}+b^{1/3})\). Then \(\frac{a^{1/3}b^{1/3}(a^{1/3}+b^{1/3})}{a^{1/3}b^{1/3}}=a^{1/3}+b^{1/3}\), which is not equal to the original expression. Wait, I think I made a mistake in the first approach. Let's do it again.

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

Let's get a common denominator of \(a^{1/3}b^{1/3}\):

\[ \frac{a^{2/3}\times a^{1/3}}{b^{1/3}\times a^{1/3}}+\frac{b^{2/3}\times b^{1/3}}{a^{1/3}\times b^{1/3}}=\frac{a^{2/3 + 1/3}+b^{2/3+1/3}}{a^{1/3}b^{1/3}}=\frac{a + b}{a^{1/3}b^{1/3}} \]

But let's check option A: \(\frac{a + b}{a^{1/3}b^{1/3}}\), option E: \(\frac{a^{2/3}b^{1/3}+a^{1/3}b^{2/3}}{a^{1/3}b^{1/3}}\). Wait, \(a^{2/3}b^{1/3}+a^{1/3}b^{2/3}=a^{1/3}b^{1/3}(a^{1/3}+b^{1/3})\), so \(\frac{a^{2/3}b^{1/3}+a^{1/3}b^{2/3}}{a^{1/3}b^{1/3}}=a^{1/3}+b^{1/3}\), which is not equal to the original expression. Wait, no, my first calculation was wrong. Let's expand the original expression:

\[ \frac{a^{2/3}}{b^{1/3}}+\frac{b^{2/3}}{a^{1/3}}=\frac{a^{2/3}\times a^{1/3}+b^{2/3}\times b^{1/3}}{a^{1/3}b^{1/3}}=\frac{a^{2/3 + 1/3}+b^{2/3+1/3}}{a^{1/3}b^{1/3}}=\frac{a + b}{a^{1/3}b^{1/3}} \]

But let's check the options. Option A is \(\frac{a + b}{a^{1/3}b^{1/3}}\), option E is \(\frac{a^{2/3}b^{1/3}+a^{1/3}b^{2/3}}{a^{1/3}b^{1/3}}\). Wait, \(a^{2/3}b^{1/3}+a^{1/3}b^{2/3}=a^{1/3}b^{1/3}(a^{1/3}+b^{1/3})\), so \(\frac{a^{2/3}b^{1/3}+a^{1/3}b^{2/3}}{a^{1/3}b^{1/3}}=a^{1/3}+b^{1/3}\), which is not equal to the original expression. Wait, I think I messed up the exponent addition. Wait, \(a^{2/3}\times a^{1/3}=a^{(2 + 1)/3}=a^{1}\), and \(b^{2/3}\times b^{1/3}=b^{1}\), so the numerator is \(a + b\) and the denominator is \(a^{1/3}b^{1/3}\), which is option A. But let's check with actual values. Let \(a = 8\) and \(b = 27\).

Original expression: \(\frac{8^{2/3}}{27^{1/3}}+\frac{27^{2/3}}{8^{1/3}}\)

\(8^{2/3}=(8^{1/3})^2 = 2^2 = 4\), \(27^{1/3}=3\), \(27^{2/3}=(27^{1/3})^2 = 3^2 = 9\), \(8^{1/3}=2\)

So original expression: \(\frac{4}{3}+\frac{9}{2}=\frac{8 + 27}{6}=\frac{35}{6}\)

Now check option A: \(\frac{8 + 27}{8^{1/3}27^{1/3}}=\frac{35}{2\times3}=\frac{35}{6}\), which matches.

Option E: \(\frac{8^{2/3}\times27^{1/3}+8^{1/3}\times27^{2/3}}{8^{1/3}27^{1/3}}\)

\(8^{2/3}\times27^{1/3}=4\times3 = 12\), \(8^{1/3}\times27^{2/3}=2\times9 = 18\), numerator: \(12 + 18 = 30\), denominator: \(2\times3 = 6\), \(\frac{30}{6}=5 eq\frac{35}{6}\)

Option A: \(\frac{8 + 27}{2\times3}=\frac{35}{6}\), which matches. Wait, but earlier I thought option E was different. So the correct answer is A? Wait, no, wait my calculation for option E was wrong. Wait, no, when \(a = 8\) and \(b = 27\), option E's numerator is \(a^{2/3}b^{1/3}+a^{1/3}b^{2/3}=8^{2/3}\times27^{1/3}+8^{1/3}\times27^{2/3}=4\times3+2\times9 = 12 + 18 = 30\), denominator is \(8^{1/3}\times27^{1/3}=2\times3 = 6\), so \(\frac{30}{6}=5\), while the original expression is \(\frac{4}{3}+\frac{9}{2}=\frac{8 + 27}{6}=\frac{35}{6}\), so option A is correct. Wait, but let's re - express the original expression:

\[ \frac{a^{2/3}}{b^{1/3}}+\frac{b^{2/3}}{a^{1/3}}=\frac{a^{2/3}\times a^{1/3}+b^{2/3}\times b^{1/3}}{a^{1/3}b^{1/3}}=\frac{a^{1}+b^{1}}{a^{1/3}b^{1/3}}=\frac{a + b}{a^{1/3}b^{1/3}} \]

So the correct option is A.

Answer:

A. \(\frac{a + b}{a^{1/3}b^{1/3}}\)