## HorseCrazyGirlForever 2 years ago Simplify the following roots. Please click on my question to see the problems because I have to draw it out. Thanks.

1. HorseCrazyGirlForever

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2. zepdrix

$\huge \sqrt[3]{16}+5\;\sqrt[3]{54}$Hmm so this is a bit tricky. We want to simplify the roots. So we need to find FACTORS of those numbers that are PERFECT CUBES. Examples of Perfect Cubes -> 8, 27, 64, ...$\huge 2 \cdot2\cdot2=8 \qquad \qquad 2^3=8$

3. zepdrix

Since neither of our numbers is larger than 64, we'll be trying to find factors of 27 or 8 under the roots.

4. zepdrix

$\huge \sqrt[3]{16} \quad \rightarrow \sqrt[3]{\color{orangered}{2}\cdot \color{orangered}{8}}$Understand how I did the first part? :D

5. HorseCrazyGirlForever

Yes.

6. zepdrix

$\huge \sqrt[3]{\color{orangered}{2}\cdot\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot \sqrt[3]{\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot 2$

7. zepdrix

Have any ideas for the second one? :)

8. HorseCrazyGirlForever

So you are trying to break it down as much as possible?

9. zepdrix

Not AS MUCH AS POSSIBLE. We certainly could have broken the 8 down a little further, into 4*2, and then into another 2*2*2.. But that wouldn't have helped us. We're trying to break down the number into a perfect cube. 8 is a perfect cube, so we were able to break it down to 8 nicely.

10. HorseCrazyGirlForever

Ohhh I see! So the factor of 54 would be 27?

11. zepdrix

Yah I think that works! 27*2 = 54

12. HorseCrazyGirlForever

Awesomeness!

13. HorseCrazyGirlForever

So it would just be written like that?

14. zepdrix

$\huge \sqrt[3]{54}\qquad =\qquad \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\qquad = \qquad \sqrt[3]{2}\cdot 3$That part make sense? We have a little bit more work to do on this one.

15. HorseCrazyGirlForever

I understand the first part, but I don't under stand how 2*27 turns into 2*3...

16. zepdrix

$\huge \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}$From here, remember that you can split up the root if you want. So we'll write it like so,$\huge \sqrt[3]{\color{royalblue}{2}}\cdot \sqrt[3]{\color{royalblue}{27}}$ From here, we just need to remember that 27 is a perfect square.$\huge 3^3=27$

17. zepdrix

So taking the cube root of 27 should give us 3.

18. HorseCrazyGirlForever

Oh okay.

19. zepdrix

So our problem has simplified to this,$\huge \sqrt[3]{16}+5\;\sqrt[3]{54} \qquad \rightarrow \qquad 2\sqrt[3]{2}+5(3\sqrt[3]{2})$

20. HorseCrazyGirlForever

Okay I understand all of that. So how would I solve the rest?

21. zepdrix

Let's multiply the 3 and 5 out.$\huge 2\sqrt[3]{2}+5(3\sqrt[3]{2}) \qquad =\qquad 2\;\sqrt[3]{2}+15\;\sqrt[3]{2}$ From here, you might notice that we have similar terms. Sure, they're kinda ugly but they should combine nonetheless.$\huge 17\; \sqrt[3]{2}$

22. zepdrix

If this part is confusing you, maybe think of the cuberoot 2 as something simpler, like x.$\huge 2x+15x=17x$

23. HorseCrazyGirlForever

Awesome! Thank you so much! I have a few more, could you please help me with those to?

24. zepdrix

Close this thread, open a new one. In the description somewhere, type @zepdrix It will give me a little popup, so I can easily find your question. There are lots of good helpers on here. So if I can't get to your question I'm sure someone can. But yah I'll come take a look ^^ heh

25. HorseCrazyGirlForever

Okay! Thank you so much!