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## HorseCrazyGirlForever Group Title Simplify the following roots. Please click on my question to see the problems because I have to draw it out. Thanks. one year ago one year ago

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1. HorseCrazyGirlForever Group Title

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2. zepdrix Group Title

$\huge \sqrt[3]{16}+5\;\sqrt[3]{54}$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 Group Title

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 Group Title

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

5. HorseCrazyGirlForever Group Title

Yes.

6. zepdrix Group Title

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

7. zepdrix Group Title

Have any ideas for the second one? :)

8. HorseCrazyGirlForever Group Title

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

9. zepdrix Group Title

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 Group Title

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

11. zepdrix Group Title

Yah I think that works! 27*2 = 54

12. HorseCrazyGirlForever Group Title

Awesomeness!

13. HorseCrazyGirlForever Group Title

So it would just be written like that?

14. zepdrix Group Title

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

15. HorseCrazyGirlForever Group Title

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

16. zepdrix Group Title

$\huge \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}$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 Group Title

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

18. HorseCrazyGirlForever Group Title

Oh okay.

19. zepdrix Group Title

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 Group Title

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

21. zepdrix Group Title

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 Group Title

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

23. HorseCrazyGirlForever Group Title

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

24. zepdrix Group Title

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 Group Title

Okay! Thank you so much!