16x^16-225y^82
Esli,
First, the area of a rectangle is length (l) time width (w). Now its not very useful to use l and w as the variables, but recognize that we can always swap l and w, that one will normally be larger and the other smaller. So lets assume for the moment that the length is longer and we will represent it as a+b and that the width is shorter and we will represent it as a-b, with a and b as positive numbers so we now have
l=a+b and w=a-b
So the area of the rectangle is now (a+b)(a-b)=a^2-b^2
So if the length times width is 16x^16-225y^82 this area must also be equal to a^2-b^2
Now 16x^16 has to be a positive number (any number raised to an even integer is positive) and
-225y^82 must be a negative number for the same reason. Similarly only a^2 can be positive and -b^2 has to be negative. So a^2=16x^16 and b^2=225y^82. Taking the square root of both sides of both equations we get
a=4x^8 and b=15y^41 and because these are real world numbers and we defined them as positive numbers, they must both be the positive roots.
So the length = a + b = 4x^8+15y^41
and the width = a - b = 4x^8-15y^41
and the width = a - b = 4x^8-15y^41
This is another use of the difference of two squares, and any time you see a pattern of term^2-term^2, it is almost always the case that using (term1-term2)(term1+term2) as part of the solution is a good idea.