- Thread starter RIO
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- Joined
- Nov 12, 2017

- Messages
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Since HCF(P, Q) = h, we can let P = xh and Q = yh, where x and y are relatively prime.

Then k = LCM(P^2, Q^3) = LCM(x^2 h^2, y^3 h^3) = x^2 y^3 h^3 . (The reason for this needs to be explained, but it is correct.)

But then P^2 Q^3 = x^2 h^2 y^3 h^3= h^2(x^2 y^3 h^3) = h^2 k

Apart from the omission of an explicit statement that x and y are relatively prime, which is necessary for my second line, you have done well.