# HCF and LCM

#### RIO

##### New member
If HCF of P and Q is h and LCM of P^2 and Q^3 is k then show that P^2Q^3=h^2k

#### Dr.Peterson

##### Elite Member
What have you tried? Where are you stuck? Please let us know, so we can have an idea where you need help. (See here.)

One approach you might try is to use the definitions of HCF and LCM. Another might involve the fact that HCF(a,b) * LCM(a,b) = ab. Are you familiar with that?

#### RIO

##### New member
Let the numbers xh,yh
LCM of P^2 and Q^3
I.e. x^2h^2 and y^3h^3= x^2y^3h^3=k
Now,
LHS-. P^2Q^3
=x^2h^2 y^3h^3
= h^2. x^2y^3h^3
= h^2 k =RHS

#### Dr.Peterson

##### Elite Member
I'll rephrase what you are saying, to make it clearer to me:

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.