Question 348066
A common factor of both expressions would probably be w^3*x^6


Your first expression would be equivalent to (w^3*x^6)*(3*w*y) = 3*w^4*x^6*y


Your second expression would be equivalent to (w^3*x^6)*(16) = 16*w^3*x^6


To confirm this is true, then use some random numbers for w and x and y and see if the relationships are equivalent.


Let w = 3 and x = 4 and y = 5.


(w^3*x^6) = 110592


(w^3*x^6) * (3*w*y) becomes 110592 * 45 = 4976640


(w^3*x^6) * (16) becomes 110592 * 16 = 1769472


3*w^4*x^6*y = 4976640 which makes it equivalent to (w^3*x^6) * (3*w*y)


16*w^3*x^6 = 1769472 which makes it equivalent to (w^3*x^6) * (16)


The common multiple appears to be:


w^3*x^6


I couldn't find any other common multiples in these equations.


The applicable basic laws of multiplication and exponentiation appear to be:


a*(b*c) = (a*b)*c


a*b = b*a


a^b * a^c = a^(b+c)