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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 is $$3$$, 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.

$$2^4 - 1^4 + 3^4 - 2^4 + 4^4 - 3^4 + \cdots + n^4 - (n - 1)^4 + (n + 1)^4 - n^4 = 4(1)^3 + 6(1)^2 + 4(1) + 1 + 4(2)^3 + 6(2)^2 + 4(2) + 1 + 4(3)^3 + 6(3)^2 + 4(3) + 1 + \cdots + 4(n - 1)^3 + 6(n - 1)^2 + 4(n - 1) + 1 + 4(n)^3 + 6(n)^2 + 4(n) + 1$$

$$2^4 + 3^4 + 4^4 + \cdots + n^4 + (n + 1)^4 - (1^4 + 2^4 + 3^4 + \cdots + (n - 1)^4 + n^4) = 4(1^3 + 2^3 + 3^3 + … + (n - 1)^3 + n^3) + 6(1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 + n^2) + 4(1 + 2 + 3 + \cdots (n - 1) + n) + (1 + 1 + 1 + ... n \ times )$$

By cancelling the same terms with opposite signs on the 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 $$1^2 + 2^2 + 3^2 + .. (n - 1)^2 + n^2$$.

${n}^{4}+4{n}^{3}+6{n}^{2}+4n=4\left({1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}\right)+6\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)+4\left(\frac{n\left(n+1\right)}{2}\right)+n$

${n}^{4}+4{n}^{3}+6{n}^{2}+4n-6\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)-4\left(\frac{n\left(n+1\right)}{2}\right)-n=4\left({1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}\right)$

${n}^{4}+4{n}^{3}+6{n}^{2}+3n-6\left(\frac{2{n}^{3}+3{n}^{2}+n}{6}\right)-4\left(\frac{{n}^{2}+n}{2}\right)=4\left({1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}\right)$

${n}^{4}+4{n}^{3}+6{n}^{2}+3n-\left(2{n}^{3}+3{n}^{2}+n\right)-2\left({n}^{2}+n\right)=4\left({1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}\right)$

${n}^{4}+4{n}^{3}+6{n}^{2}+3n-2{n}^{3}-3{n}^{2}-n-2{n}^{2}-2n=4\left({1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}\right)$

${n}^{4}+2{n}^{3}+{n}^{2}=4\left({1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}\right)$

$\frac{{n}^{4}+2{n}^{3}+{n}^{2}}{4}={1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}$

$\frac{{n}^{2}\left({n}^{2}+2n+1\right)}{4}={1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}$

$\frac{{n}^{2}{\left(n+1\right)}^{2}}{4}={1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}$

Therefore, ${1}^{3}+{2}^{3}+{3}^{3}+...{\left(n-1\right)}^{3}+{n}^{3}={\left(\frac{n\left(n+1\right)}{2}\right)}^{2}$
Sum of cubes of first $$n$$ natural numbers $$=$$ ${\left(\frac{n\left(n+1\right)}{2}\right)}^{2}$
Important!
Amicable numbers, or friendly numbers, are a pair of numbers whose sum of proper divisors equals the other.