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The number $(\sqrt{2}+\sqrt{3})^3$ can be written in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$, where $a$, $b$, and $c$ are integers. What is $a+b+c$? 

 

The number \((\sqrt{2}+\sqrt{3})^3\) can be written in the form  \(a\sqrt{2} + b\sqrt{3} + c\sqrt{6}\), where \(a\), \(b\), and \(c\) are integers. What is \(a+b+c\)

 

The number (√2+√3)³ can be written in the form a√2+b√3+c√6, where a, b, and c are integers. What is a+b+c? 

 

 

i put three forms of the same question

THX EVERYBODY!

 Dec 8, 2018
 #1
avatar+99331 
+2

The number (√2+√3)³ can be written in the form a√2+b√3+c√6, where a, b, and c are integers. What is a+b+c? 

 

\((\sqrt2+\sqrt3)^3=(\sqrt3)^3+3(\sqrt3)^2(\sqrt2)+3(\sqrt3)(\sqrt2)^2+(\sqrt2)^3\\ \)

 

Now expand it out and get the answer for yourself.  If you have new problems with this question then let us know.

 Dec 8, 2018
 #3
avatar+15 
+2

I got \(11\sqrt2+9\sqrt3\), but how would I turn this into the form of  \(a\sqrt{2} + b\sqrt{3} + c\sqrt{6}\)

THX FOR EVERYBODY'S HELP!

qpwoei  Dec 9, 2018
 #5
avatar+99331 
+2

I got a different answer from you but I did not get any sqrt6   

so I suppose the c is 0

Melody  Dec 9, 2018
 #2
avatar+3994 
+2

I can clearly read the second one. Thanks!

Hint: Apply sum of cubes formula.

 Dec 8, 2018
 #4
avatar+99331 
+2

Yes thanks for posting clearly.   laugh

 

Here is the version I would have liked best 

 

\(\text{The number }\;(\sqrt{2}+\sqrt{3})^3 \text{ can be written in the form }\\a\sqrt{2} + b\sqrt{3} + c\sqrt{6}, \text{ where $a$, $b$, and care integers. What is a+b+c ?}\)

 

 

I like this version best becasue it is the easiest to copy and to work on.     

 Dec 9, 2018
 #6
avatar+3994 
+2

I'll post a solution:

We just apply the sum of cubes formula, so, we get \(\left(\sqrt{2}\right)^3+3\left(\sqrt{2}\right)^2\sqrt{3}+3\sqrt{2}\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3\)

 

We just simplify this and add like terms, to reach, \(11\sqrt{2}+9\sqrt{3}.\)

 

From here, we know that \(a=11, b=9, c=0\)  , thus  \(11+9+0=\boxed{20}.\)

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 Dec 9, 2018
edited by tertre  Dec 9, 2018
 #8
avatar+15 
+2

\(\textrm{OH! I didn't think of c being 0. Tricky! (P.S. a=11, b=9, and c=0, not b=6)}\)

qpwoei  Dec 9, 2018
 #7
avatar+15 
+2

\(\textrm{OK, got it! So, I still don't understand how I would turn }11\sqrt2+9\sqrt3\textrm{ into the form of }a\sqrt{2} + b\sqrt{3} + c\sqrt{6}. \)

 

EDIT: nevermind, I got it. THANK YOU TO EVERYBODY FOR HELPING ME WITH THIS PROBLEM! 

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 Dec 9, 2018
edited by Guest  Dec 9, 2018

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