Sum of cubes of even and odd natural numbers

We know that sum of cubes of first n natural numbers is = (n(n+1)/2)². 

Sum of cube of first n even natural numbers 2³ + 4³ + 6³ + ……… + (2n)³

Even Sum = 23 + 43 + 63 + .... + (2n)3
        if we multiply by 23 then 
        = 23 x (13 + 23 + 33 + .... + (n)3)
        = 23 + 43 + 63 + ......... + (2n)3
        =  23 (n(n+1)/2)2
        =  8(n(n+1))2/4
        =  2(n(n+1))2

Example : 

Sum of cube of first 4 even numbers = 23 + 43 + 63 + 83 
 put n = 4     = 2(n(n+1))2
               = 2*(4*(4+1))2
               = 2(4*5)2
               = 2(20)2
               = 800
 8 + 64 + 256 + 512 = 800

Program for Sum of cubes of first n even numbers

Sum of cube of first n odd natural numbers We need to compute 1³ + 3³ + 5³ + …. + (2n-1)³

OddSum   = (Sum of cubes of all 2n numbers) - (Sum of cubes of first n even numbers)
         =  (2n(2n+1)/2)2 - 2(n(n+1))2 
         =  n2(2n+1)2 - 2* n2(n+1)2
         =  n2[(2n+1)2 - 2*(n+1)2]
         =  n2[4n2 + 1 + 4n - 2n2 - 2 - 4n]
         =  n2(2n2 - 1)

Example : 

Sum of cube of first 4 odd numbers = 13 + 33 + 53 + 73 
 put n = 4     = n2(2n2 - 1)
               = 42(2*(4)2 - 1)
               = 16(32-1)
               =  496
 1 + 27 + 125 + 343 = 496

Program for Sum of cubes of first n odd numbers

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