Question

If y>x>1, and (x^1/3 y^1/6) = 256 what is the value of xy?

Answer 1

Answer:

xy = 2^39.

Step-by-step explanation:

8 *32 = 256

so x could be 8^3 and y could be 32^6

So xy = 8^3 * 32^6

= (2^3)^3 * (2^5)^6

= 2^9 * 2^30

= 2^39.