Sums of Cubes
The following contains Sums of 4 Cubes, 5 Cubes, etc. that can be expressed in 2 ways, 3 ways, etc. Click here for a Graphics Version of this page
Other Links : PELL NUMBERS FORMULAE , SUMS OF POWERS, FIBONACCI FORMULAE, RAJESH's MATHS PAGE, TRIANGLE NUMBERS THAT ARE PERFECT SQUARES
Also see Sums of Cubes , Sums of Powers - Ramanujan and his Number 1729
| (ax + by + ay)^3 + (bx)^3 + (ax + by + bx)^3 +
(ay)^3 = (ax + ay + bx)^3 + (by)^3 + (ay + by + bx)^3 + (ax)^3 |
SOP3_1 |
| (ax + ay + bx)^3 + (ax + ay + by)^3 + (ax + bx +
by)^3 + (ax - by)^3 = (ax)^3 + (ax + bx)^3 + (ax + ay)^3 + (ax + bx + ay + by)^3 |
SOP3_2 |
| (ax + by + ay)^3 + (bx)^3 + (-ax - ay - bx)^3 +
(-by)^3 = (-ax - by - bx)^3 + (-ay)^3 + (ay + by + bx)^3 + (ax)^3 = (ay + by)^3 + (-ax - bx)^3 + (ax - by)^3 + (bx - ay)^3 |
SOP3_3 |
| This equation below is actually an extension of one given by
Ramanujan for Sum of Two Cubes in two ways. IF a^2 + ab + b^2 = 3cd^2 THEN (dc^2 + a)^3 - (ac + d)^3 = (dc^2 + b)^3 - (bc + d)^3 = (dc^2 - a - b)^3 + (ac + bc - d)^3 (dc^2 - a)^3 + (ac - d)^3 = (dc^2 - b)^3 + (bc - d)^3 = (dc^2 + a + b)^3 - (ac + bc + d)^3 |
SOP3_4 |
| 3 ( (a^2 + 2ab)^3 + (-b^2 - 2ab)^3 + (b^2 - a^2)^3
) = ( (-a)^3 + (-b)^3 + (a + b)^3 ) * ( (-a - 2b)^3 + (2a + b)^3 + (b - a)^3 ) |
SOP3_5 |