LCM of a4b3, a3b6 and a4b8 is axby. Value of yx =
The L.C.M. of a4b3, a3b6 and a4b8 is a4b8.
∴x=4 and y=8
So, yx = 2
The LCM of x2y + xy2 and x2 + xy is
If a and b are different primes such that (i)[a−1b2a2b−4]7÷[a3b−5a−2b3]=axby,find x and y.
(ii)(a+b)−1(a−1+b−1)=axby,find x+y+2.