# Properties of Roots of Real Numbers

1. Jan 16, 2005

### IndigoSwing4

I have three problems that I can't seem to solve. I was wondering if anybody could help me or explain to me how to solve these. Note: * = multiplication.

1. (6^1/2 * 2^1/3)^6

2. ^4√7 + 2^4√1792

3. 3(X-4)^1/2 + 5 = 11

2. Jan 16, 2005

### robert Ihnot

#2 seems to be missing a term. What is raised to 4 times square root of 7?

#3 seems easy, I think. It reduces to: $$\sqrt{x-4}=2.$$
That gives x= 8, which works.

For #1, use the binominal expansion: $$(a+b)^6= a^6+6a^5b+15a^4b^2 +20a^3b^3+15a^2b^4+6ab^5+b^6.$$

Last edited: Jan 16, 2005
3. Jan 17, 2005

### HallsofIvy

Staff Emeritus
1. (6^(1/2)*2^(1/3))^6

That's "*", not "+", so you don't need the formula robert Ihnot gave:

(6^(1/2)*2^(1/3))^6= (6^(1/2))^6*(2^(1/3))^6
= 6^((1/2)*6)*2^((1/3)*6)= 6^3*2^2= 216*4= 864.