If x = 1, y = 2 and z = 3, the expression becomes a perfect square.

So, the LCM is 6


Regards
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I quickly thought it was 1*2*3 = 6 in 10second because x,y,z are differentiated and >0, therefore the "most" or "only" possible positive integers that I could think of were 1,2 and 3.
Hopefully my thinking was on the right track for this problem

I thought like this 24 = 3*4*2 so if x=3, y=4 & z= 2 then it becomes 3^4*4^4*2^4 which is a square number. so the lcm will be 12. Why it can't be ?

Basically

\(3^4 \times 4^4 \times 2^4\)

Is \(3^4 \times 2^{12}\)

so the LCM is \(3 \times 2 = 6\)

Regards
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I started by breaking down 24 to its prime, which is (2)^3 and (3)^1. Then I got stuck and did not know what to do next. Help?!

Let \(A^2 = 24x^3y^3z^3\)

\(A = \sqrt{24x^3y^3z^3}\)

\(A = \sqrt{(2)(2)(2)(3)x^3y^3z^3}\)

\(A = 2xyz \sqrt{(2)(3)xyz}\)

Notice, the product of \(xyz\) can only be 6
So, their LCM = 6
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