Ramanujan’s Identities (with source code)
Author: hidekichi45Executions: 302
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Ramanujan's Identities (with source code) Last updated February 28, 2022
(6*a^2-4*a*b+4*b^2)^3=(3*a^2+5*a*b-5*b^2)^3+(4*a^2-4*a*b+6*b^2)^3+(5*a^2-5*a*b-3*b^2)^3 (A)
or
(6*a^2-4*a*b+4*b^2)^3-(5*a^2-5*a*b-3*b^2)^3=(3*a^2+5*a*b-5*b^2)^3+(4*a^2-4*a*b+6*b^2)^3
=(6*a^2-4*a*b+4*b^2)^3+(-5*a^2+5*a*b+3*b^2)^3=(3*a^2+5*a*b - 5*b^2)^3 + (4*a^2 - 4*a*b + 6*b^2)^3 (B)
Let’s verify this by actually substituting values. Since the calculation has a precision of 50 digits, please limit your input to 7 digits.
If you use a 9-digit number, a digit overflow will occur, resulting in calculation errors.
or
(6*a^2-4*a*b+4*b^2)^3-(5*a^2-5*a*b-3*b^2)^3=(3*a^2+5*a*b-5*b^2)^3+(4*a^2-4*a*b+6*b^2)^3
=(6*a^2-4*a*b+4*b^2)^3+(-5*a^2+5*a*b+3*b^2)^3=(3*a^2+5*a*b - 5*b^2)^3 + (4*a^2 - 4*a*b + 6*b^2)^3 (B)
Let’s verify this by actually substituting values. Since the calculation has a precision of 50 digits, please limit your input to 7 digits.
If you use a 9-digit number, a digit overflow will occur, resulting in calculation errors.
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