any two cubic number that when added toghether gives anothere cubic number a^3+b^3=c^3. find a,b&c
No, your premise is not true.
2^3+3^2= a^3
8+27=
35, and 35 is not a cube of a natural number.
Fermat's Last Theorem states that there are no solutions to x^n + y^n = z^n for n = 3 or greater. Surprisingly, there are solutions to x^n + y^n = z^(n+1) or z^(n-1).
1--Divide through by z^n resulting in (x/z)^n + (y/z)^n = z.
2--Let (x/z) and (y/z) be any positive numbers other than 1.
3--Solve for z and then x and y.
4--Example: x^3 + y^3 = z^4
5--Let (x/z) = 2 and (y/z) = 5. Then 2^3 + 5^3 = 133 = z.
6--Then x = 2(133) = 266 and y = 5(133) = 665.
7--Checking: 266^3 + 665^3 = 18,821,096 + 294,079,625 = 312,900,721 = 133^4.
It works for any n and (n+1).
To find two cubic numbers, a^3 and b^3, that sum to give another cubic number, c^3, we need to look for a special case known as a "cubic sum."
A cubic sum occurs when the sum of two cubes can be expressed as another cube. In this case, we are looking for solutions to the equation a^3 + b^3 = c^3.
One such example of a cubic sum is the famous case of Fermat's Last Theorem, which states that for any whole numbers a, b, and c, the equation a^n + b^n = c^n has no solutions when n is a positive integer greater than 2.
Therefore, in the case of a^3 + b^3 = c^3, we can conclude that there are no whole number solutions for this equation.
However, if you are looking for non-whole number solutions (such as fractions or decimals), you may need to use numerical methods or approximation techniques to find suitable values for a, b, and c.