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Theory:

The natural numbers are \(1\), \(2\), \(3\), \(4\), …
 
We need to find value of \(1^3 + 2^3 + 3^3  + …. + n^3\).
 
Consider the identity \((x + 1)^{k + 1} - x^{k + 1}\).
 
Since the power of all the numbers are \(3\), put \(k = 3\) in the above identity.
 
\((x + 1)^{3 + 1} - x^{3 + 1} = (x + 1)^4 - x^4\)
 
\(= ((x + 1)^2)^2 - x^4\)
 
\(= (x^2 + 2x + 1^2)^2 - x^4\)
 
\(= x^4 + 4x^2 + 1^4 + 4x^3 + 4x + 2x^2 - x^4\)
 
\(= 4x^3 + 6x^2 + 4x + 1\)
 
\((x + 1)^4 - x^4 = 4x^3 + 6x^2 + 4x + 1\) - - - - (I)
 
Now, substitute \(x = 1, 2, 3, … n\) in equation (I).
 
When \(x = 1\), \(2^4 - 1^4 = 4(1)^3 + 6(1)^2 + 4(1) + 1\)
 
When \(x = 2\), \(3^4 - 2^4 = 4(2)^3 + 6(2)^2 + 4(2) + 1\)
 
When \(x = 3\), \(4^4 - 3^4 = 4(3)^3 + 6(3)^2 + 4(3) + 1\)
 
           \(\vdots\)             \(\vdots\)             \(\vdots\)
 
When \(x = n - 1\), \(n^4 - (n - 1)^4 = 4(n - 1)^3 + 6(n - 1)^2 + 4(n - 1) + 1\)
 
When \(x = n\), \((n + 1)^4 - n^4 = 4(n)^3 + 6(n)^2 + 4(n) + 1\)
 
Add all the above equations of \(x\) values.
 
\(2^4 - 1^4 + 3^4 - 2^4 + 4^4 - 3^4  + \cdots + n^4 - (n - 1)^4 + (n + 1)^4 - n^4 = 4(1)^3 + 6(1)^2 + 4(1) + 1 + 4(2)^3 + 6(2)^2 + 4(2) + 1 + 4(3)^3 + 6(3)^2 + 4(3) + 1 + \cdots + 4(n - 1)^3 + 6(n - 1)^2 + 4(n - 1) + 1 + 4(n)^3 + 6(n)^2 + 4(n) + 1\)
 
\(2^4 + 3^4 + 4^4 + \cdots + n^4 + (n + 1)^4 - (1^4 + 2^4 + 3^4 + \cdots + (n - 1)^4 + n^4) = 4(1^3 + 2^3 + 3^3 + … + (n - 1)^3 + n^3) + 6(1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 + n^2) + 4(1 + 2 + 3 + \cdots (n - 1) + n) + (1 + 1 + 1 + ... n \ times )\)
 
Cancelling the same terms with opposite signs on LHS, we get:
 
\((n + 1)^4 - 1^4 = 4(1^3 + 2^3 + 3^3 + … + (n - 1)^3 + n^3) + 6(1^2 + 2^2 + 3^2 + … (n - 1)^2 + n^2) + 4(1 + 2 + 3 + … (n - 1) + n) + n\)
 
\(n^4 + 4n^3 + 6n^2 + 4n + 1^4 - 1^4 = 4(1^3 + 2^3 + 3^3 + … + (n - 1)^3 + n^3) + 6(1^2 + 2^2 + 3^2 + … (n - 1)^2 + n^2) + 4(1 + 2 + 3 + … (n - 1) + n) + n\)
 
Apply sum of first \(n\) terms of natural numbers formula in \(1 + 2 + 3 + .. (n - 1) + n\).
 
And, apply sum of squares of first \(n\) terms of natural numbers formula in \(1^2 + 2^2 + 3^2 + .. (n - 1)^2 + n^2\).
 
n4+4n3+6n2+4n=4(13+23+33+...(n1)3+n3)+6n(n+1)(2n+1)6+4n(n+1)2+n
 
n4+4n3+6n2+4n6n(n+1)(2n+1)64n(n+1)2n=4(13+23+33+...(n1)3+n3)
 
n4+4n3+6n2+3n62n3+3n2+n64n2+n2=4(13+23+33+...(n1)3+n3)
 
n4+4n3+6n2+3n(2n3+3n2+n)2(n2+n)=4(13+23+33+...(n1)3+n3)
 
n4+4n3+6n2+3n2n33n2n2n22n=4(13+23+33+...(n1)3+n3)
 
n4+2n3+n2=4(13+23+33+...(n1)3+n3)
 
n4+2n3+n24=13+23+33+...(n1)3+n3
 
n2n2+2n+14=13+23+33+...(n1)3+n3
 
n2n+124=13+23+33+...(n1)3+n3
 
Therefore, 13+23+33+...(n1)3+n3=n(n+1)22
Sum of cubes of first \(n\) natural numbers \(=\) n(n+1)22
Important!
Amicable numbers or friendly numbers are a pair of numbers whose sum of proper divisors equals the other.