Question

Rationalize the Denominator and simplify.

Answer 1

\[\sqrt[3]{\frac{ y }{ 4 }}+\sqrt[3]{\frac{ y }{ 32 }}-\sqrt[3]{\frac{ y }{ 108 }}\]
So far I have
\[\sqrt[3]{\frac{ 2y }{ 2 }}+\sqrt[3]{\frac{ 2y }{ 4 }}-\sqrt[3]{\frac{ 2y }{ 6 }}\]

Answer 2

The first step is to factor each denominator into prime factors to see what you need to multiply each denominator by to get rid of each cubic root in the denominator.

\(\sqrt[3]{\dfrac{ y }{ 4 }}+\sqrt[3]{\dfrac{ y }{ 32 }}-\sqrt[3]{\dfrac{ y }{ 108 }} =\)

\(=\sqrt[3]{\dfrac{ y }{ 2^2 }}+\sqrt[3]{\dfrac{ y }{ 2^5 }}-\sqrt[3]{\dfrac{ y }{ 2^2 \times 3^3 }} \)

Each denominator has to have a base raised to a power that a multiple of 3, so you can take the cubic root.

Answer 3

\(=\sqrt[3]{\dfrac{ y }{ 2^2 }} \times \dfrac{\sqrt[3]2}{\sqrt[3] 2}+\sqrt[3]{\dfrac{ y }{ 2^5 }} \times \dfrac{\sqrt[3]2}{\sqrt[3] 2} -\sqrt[3]{\dfrac{ y }{ 2^2 \times 3^3 }} \times \dfrac{\sqrt[3]2}{\sqrt[3] 2}\)

\(=\sqrt[3]{\dfrac{ 2y }{ 2^3} } +\sqrt[3]{\dfrac{ 2y }{ 2^6 }} -\sqrt[3]{\dfrac{ 2y }{ 2^3 \times 3^3 }} \)

Now we separate the cubic roots of the numerators and denominators.

\(=\dfrac{\sqrt[3]{ 2y }} { \sqrt[3]{2^3}} +\dfrac{\sqrt[3]{ 2y }}{\sqrt[3]{ 2^6 }} -
\dfrac{\sqrt[3]{ 2y }}{ \sqrt[3]{2^3 }} \)

Since each cubic root in a denominator is the cubic root of a cube, we take the roots in the denominators.

\(=\dfrac{\sqrt[3]{ 2y }} {2} +\dfrac{\sqrt[3]{ 2y }}{2^2 } -
\dfrac{\sqrt[3]{ 2y }}{ 2} \)

\(=\dfrac{\sqrt[3]{ 2y }} {2} +\dfrac{\sqrt[3]{ 2y }}{4 } -
\dfrac{\sqrt[3]{ 2y }}{ 2} \)

Answer 4

Now you need a common denominator to do the addition and the subtraction of fractions.

\(=\dfrac{2\sqrt[3]{ 2y }} {4} +\dfrac{\sqrt[3]{ 2y }}{4 } -
\dfrac{2\sqrt[3]{ 2y }}{ 4}\)

Answer 5

Now treat \(\sqrt[3]{2y} \) as if it were \(x\).
That makes all numerators like terms, and they can be combined together.

\(=\cancel{\dfrac{2\sqrt[3]{ 2y }} {4}} +\dfrac{\sqrt[3]{ 2y }}{4 } -
\cancel{\dfrac{2\sqrt[3]{ 2y }}{ 4}}\)

\(=\dfrac{\sqrt[3]{ 2y }}{4 }\)

Answer 6

Do you understand?

Answer 7

Kind of, yes...but I plugged it into my homework and it said it was incorrect

Answer 8

I see a mistake I made above in the third denominator.
Let me redo it from that point on.

Answer 9

Now we separate the cubic roots of the numerators and denominators.

\(=\dfrac{\sqrt[3]{ 2y }} { \sqrt[3]{2^3}} +\dfrac{\sqrt[3]{ 2y }}{\sqrt[3]{ 2^6 }} -
\dfrac{\sqrt[3]{ 2y }}{ \sqrt[3]{2^3 \times 3^3}}\)

\(=\dfrac{\sqrt[3]{ 2y }} {2} +\dfrac{\sqrt[3]{ 2y }}{4 } -
\dfrac{\sqrt[3]{ 2y }}{ 6}\)

Notice that this is similar to what you have, but there are no roots in the denominator.

Answer 10

\(=\dfrac{6\sqrt[3]{ 2y }} {12} +\dfrac{3\sqrt[3]{ 2y }}{12 } -
\dfrac{2\sqrt[3]{ 2y }}{ 12}\)

\(=\dfrac{7\sqrt[3]{ 2y }} {12} \)

Answer 11

Awesome! Thanks