adding radicals

**isundae** · September 7th 2009, 05:44 PM

I don't know what to do for this one... the exponent outside the radical confuses me:

$\sqrt[3]{16} + 3\sqrt[3]{54}$

It says to simplify the expression. I know that I have to factor the numbers, but that exponent "3" is confusing. Help?

**Prove It** · September 7th 2009, 05:47 PM

Originally Posted by isundae

I don't know what to do for this one... the exponent outside the radical confuses me:

$\sqrt[3]{16} + 3\sqrt[3]{54}$

It says to simplify the expression. I know that I have to factor the numbers, but that exponent "3" is confusing. Help?

The same rules apply as when they are square roots. The only difference is now you have to find a CUBIC factor, rather than a square factor...

Note that $16 = 8\cdot 2$

So $\sqrt[3]{16} = \sqrt[3]{8\cdot 2} = \sqrt[3]{8}\cdot\sqrt[3]{2} = 2\cdot\sqrt[3]{2}$ .

Also, note that $54 = 27\cdot 2$ .

So $\sqrt[3]{54} = \sqrt[3]{27\cdot 2} = \sqrt[3]{27}\cdot\sqrt[3]{2} = 3\cdot\sqrt[3]{2}$ .

Therefore $\sqrt[3]{16} + 3\cdot\sqrt[3]{54} = 2\cdot\sqrt[3]{2} + 9\cdot\sqrt[3]{2} = 11\cdot\sqrt[3]{2}$ .

**isundae** · September 7th 2009, 05:56 PM

Originally Posted by Prove It

The same rules apply as when they are square roots. The only difference is now you have to find a CUBIC factor, rather than a square factor...

Note that $16 = 8\cdot 2$

So $\sqrt[3]{16} = \sqrt[3]{8\cdot 2} = \sqrt[3]{8}\cdot\sqrt[3]{2} = 2\cdot\sqrt[3]{2}$ .

Also, note that $54 = 27\cdot 2$ .

So $\sqrt[3]{54} = \sqrt[3]{27\cdot 2} = \sqrt[3]{27}\cdot\sqrt[3]{2} = 3\cdot\sqrt[3]{2}$ .

Therefore $\sqrt[3]{16} + 3\cdot\sqrt[3]{54} = 2\cdot\sqrt[3]{2} + 9\cdot\sqrt[3]{2} = 11\cdot\sqrt[3]{2}$ .

Oh, I see now. I forgot that I could take the equation apart, that makes it much easier

Just wondering, where did the 9 come from? (in the last step)

edit: Never mind, I see where it came from...

Thanks for helping me!

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