Question

If two positive integers 'a' and 'b' are expressible in the form a =pq^2, b=p^2q,p,q are prime numbers then show that LCM (a,b)×HCF(a,b)=a×b.

Answer 1

We have,

$a=p q^{2}$

$b= p^{2} q$


Let us find the LCM and HCF of a and b,

 HCF is going to be pq   and LCM is going to be $p^{2} q^{2}$

now we have to prove,

LCM * HCF = PRODUCT OF THE NUMBERS

Let us first take the LHS side,

LCM * HCF = $pq* p^{2} q^{2}$
                  = $p^{3} q^{3}$
                  = $p q^{2} * p^{2} q$   (rearranging)
                  = a*b
                  = product of the numbers

HENCE PROVED......

HOPE THIS HELPS.....PLZ MARK AS BRAINLIEST
 

Answer 2

Answer:

a=pq^2

= p*q*q

b=p^3q

=p*p*p*q

therefore lcm(a,b)= p*p*p*q*q

=p^3q^2