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x-y=9 x*y=13 x² + y² = ( x - y )² + 2xy = 9² + 2(13) = 81 + 26 = 107

let the numbers be x and y we have x-y=9 and x*y=13

we know that (x-y)^2=x^2 + y^2 -2*x*y so, 81=x^2 + y^2 -2*13 x^2 + y^2= 81+26 x^2 + y^2 = 107 hence, sum of squares=107

let the numbers be x and y we have x-y=9 and x*y=13 we know that (x-y)^2=x^2 + y^2 -2*x*y so, 81=x^2 + y^2 -2*13 x^2 + y^2= 81+26 x^2 + y^2 = 107 hence, sum of squares=107

x-y=9 xy=13 (x-y)^2=x^2 +y^2-2xy 81=x^2+y^2-26 107=x^2+y^2

(x-y)^2 = x^2 - 2xy + y^2 = 81. x^2 + y ^2 = 81+ 2xy 2 xy = 2*13 = 26 So, the sum of their squares, x^2 + y^2 = 81+26 = 107.

There are two whole numbers, difference of their squares is a cube and difference of their cube is a square. These are the smallest possible numbers. can you find the numbers?