SOLUTION:
Let x and y be the two legs of each right triangle.
As shown in the figure, the area of the small square in the middle part is 1 square unit, and the area of the large square is 13 square units.
Thus, the sum of the four congruent right triangles is 13 - 1 = 12 square units.
Now, add a right triangle with same area to the outside of each right triangle such that this forms another larger square.
Hence, the area of the larger square is 13 + 12 = 25 square units.
This implies that the side of the square is 5 units as s² = Area.
Moreover, we can observe that the sum of the two legs of the right triangle is equal to the side length of the largest square, which is 5 units.
Also, the difference between the two legs is exactly equal to the side length of the original square; that is 1 unit.
So,
[tex]\begin{cases} \sf x + y = 5\\ \sf x - y = 1\\\end{cases}[/tex]
Adding the two equations and solving, we get (x,y) = (3,2).
Therefore, the sum of the cubes of the two legs of each right triangle is
3³ + 2³ = 35 units
ANSWER:
35 units.
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