Find ab given its relationship to the number 432

[fk378]

If a,b are positive integers and (a^1/2b^1/3)⁶ = 432, then what is the value of ab?

 

[jedishrfu]

If a,b are positive integers and (a^1/2b^1/3)⁶ = 432, then what is the value of ab?

Is this a problem from the SAT?

First bring the 6 inside the a and b term to get a^6/2 * b^6/3 = 432

 

[fk378]

Is this a problem from the SAT?

First bring the 6 inside the a and b term to get a^6/2 * b^6/3 = 432

Yes, I did that. Don't know where to go from here. Seems like I just go in circles when I try to make two equations to solve for the two unknowns.

 

[fk378]

Is the only way to do this just to get a³b²=432, then find the factors of 432? I tried this and then got 16*27=432, so then a=3, b=2. But I feel like there must be a different way to do this problem...

 

[L4xord]

Hmmmm... It would be good if there was another way. It seems a bit too easy.

 

[Mentallic]

Is the only way to do this just to get a³b²=432, then find the factors of 432? I tried this and then got 16*27=432, so then a=3, b=2. But I feel like there must be a different way to do this problem...

You mean b=4

I'm not aware of another way if there is one, and I'd imagine if there were, it'd be fairly more complicated.

 

[pwsnafu]

If a,b are positive integers and (a^1/2b^1/3)⁶ = 432, then what is the value of ab?

We are given
$(a^{1/2} b^{1/3})^6 = 432$
So
$a^3 b^2 = a(ab)^2 = 432$
$(ab)^2 = \frac{432}{a}$
LHS is a square, so test different a.
$a = 2 \implies \frac{432}{a} = 216$ not a square
$a = 3 \implies \frac{432}{a} = 144$
144 is a square, so ab = 12.

 

[Mentallic]

We are given
$(a^{1/2} b^{1/3})^6 = 432$
So
$a^3 b^2 = a(ab)^2 = 432$
$(ab)^2 = \frac{432}{a}$
LHS is a square, so test different a.
$a = 2 \implies \frac{432}{a} = 216$ not a square
$a = 3 \implies \frac{432}{a} = 144$
144 is a square, so ab = 12.

Nice!

 

[HallsofIvy]

Notice that the condition "a,b are positive integers" is crucial here. If a and b were allowed to be negative, there would be more solutions. If a and b were allowed to be any real numbers there would be an infinite number of solutions.

 

[pwsnafu]

Notice that the condition "a,b are positive integers" is crucial here. If a and b were allowed to be negative, there would be more solutions.

If a and b were allowed to be negative and we are allowed to use complex algebra, then yes.
Otherwise a^1/2 is undefined.