+0

# HELP THIS MATH QUESTION IS DUE: 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?

+1
250
8
+15

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
+102459
+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
+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
+102459
+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
+4296
+2

I can clearly read the second one. Thanks!

Hint: Apply sum of cubes formula.

Dec 8, 2018
#4
+102459
+2

Yes thanks for posting clearly.

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
+4296
+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  $$11+9+0=\boxed{20}.$$

.
Dec 9, 2018
edited by tertre  Dec 9, 2018
#8
+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
+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!

.
Dec 9, 2018
edited by Guest  Dec 9, 2018