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First of all, there’s a family of trivial solutions of the form[math]\displaystyle m^3+(-m)^3+10^3=10^3[/math]Those are not very exciting. A much more interesting infinite family of solutions was given by Mahler in 1936, though I’m not sure if he was the first to identify it. Observe that, for any [math]t[/math],[math]\displaystyle (9t^4)^3+(3t-9t^4)^3+(1-9t^3)^3=1[/math]This is a polynomial identity: it holds universally, as an equality of polynomials, and is therefore true for any value of [math]t[/math] whatsoever. So let [math]t[/math] be any integer [math]m[/math] and multiply the equation by [math]10^3[/math] to obtain the infinite family[math]\displaystyle (90m^4)^3+(30m-90m^4)^3+(10-90m^3)^3=10^3[/math]For example, when [math]m=1[/math], this is [math]90^3+(-60)^3+(-80)^3=10^3[/math] which is simply a rearrangement of [math]9^3=1^3+6^3+8^3[/math].See this paper by Mordell[1] for more on this. There are similar parametrizations for equations of the form [math]X^3+Y^3+Z^3=2k^3[/math].There are other, “sporadic” solutions which aren’t covered by those infinite families, for example Ramanujan’s famous taxicab identity[math]12^3+1^3+(-9)^3=10^3[/math]There are many other such “sporadic” solutions and I’m not sure if they are fully classified. Here are a few:[math]\displaystyle \begin{align} 12^3+(-6)^3+(-8)^3 &= 10^3 \\ 19^3 + (-3)^3+(-18)^3 &= 10^3 \\ 24^3 + 19^3 + (-27)^3 &= 10^3 \\ 75^3 + 45^3 + (-80)^3 &= 10^3 \end{align}[/math]Footnotes[1] Journal of the London Mathematical Society

1729 is known as the Hardy–Ramanujan number after a famous anecdote of the British mathematician G. H. Hardy regarding a visit to the hospital to see the Indian mathematician Srinivasa Ramanujan.In Hardy's words : "I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."The two different ways are these:1729 = 1^3 + 12^3 = 9^3 + 10^3The quotation is sometimes expressed using the term "positive cubes", since allowing negative perfect cubes (the cube of anegative integer) gives the smallest solution as 91 (which is a divisor of 1729):91 = 6^3 + (−5)^3 = 4^3 + 3^3Source : Wikipedia .Ramanujan's square​​​​​​​​​​​​​Source : Mathematician ramanujan's square .

Answer: 100The Solution:(1+1) ×( 1+1) = 4,(2+2) × (2+2) = 16,(3+3) × (3+3) = 36,(4+4) x (4+4)=64,so, (5+5) ×(5+5)= 100