Let $n$ be a positive integer. Prove that \[n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\]by counting the number of ordered triples $(a,b,c)$ of positive integers, where $1 \le a,$ $b,$ $c \le n,$ in two different ways.

Filliqq Nov 1, 2020

#1**0 **

Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face! And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c). If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks. There are n ways to choose a, and there are n - 1 ways to choose b. We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c. So there are 3n(n - 1) smirks.

Now we count the number of smiley faces. There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c. So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces. By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

Guest Nov 2, 2020

#2**+2 **

1 is a positive integer and if n=1 then the expression makes no sense.

So I suppose that is an instant disproof.

If it is true at all, n would have to be a positive integer greater or equal to 3

I have absolutely no idea what the last sentence in the question is referring to.

Melody Nov 2, 2020

#3**+2 **

\(n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\)

Prove by induction for n>=3

__Step 1__

Prove true for n=3

LHS =27

RHS=27

So the statement is true of n=3

Step 2

Assume true for n =k where k is some integer greater or equal to 3

So this statement is true:

\(k^3=k+3k(k-1)+6\binom{k}{3}\\ k^3=k+3k^2-3k+6*\frac{k!}{(k-3)!3!}\\ k^3=3k^2-2k+\frac{k!}{(k-3)!}\\ k^3-3k^2+2k=\frac{k!}{(k-3)!}\\ \)

NOW prove that it will also be true for n=k+1

i.e. Prove

\((k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\~\\ LHS=x^3+3k^2+3k+1\\~\\ RHS=k+1+3k^2+3k+6*\frac{(k+1)!}{3!(k-2)!}\\ RHS=3k^2+4k+1+\frac{(k+1)!}{(k-2)!}\\ RHS=3k^2+4k+1+\frac{k!(k+1)}{(k-3)!(k-2)}\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*\frac{k!}{(k-3)!}\\ sub\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*(k^3-3k^2+2k)\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k^2-3k+2)\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k-2)(k-1)\\ RHS=3k^2+4k+1+(k+1)*k(k-1)\\ RHS=3k^2+4k+1+k^3-k\\ RHS=k^3+3k^2+3k+1\\ RHS=LHS\)

So if it is true for n=k, it will be true for n=k+1

__Step 3__

The statement is true for n=3 so it is true for n=4, n=5, .....

Hence the statement is true for all integer values of n greater or equal to 3.

__LaTex coding:__

k^3=k+3k(k-1)+6\binom{k}{3}\\

k^3=k+3k^2-3k+6*\frac{k!}{(k-3)!3!}\\

k^3=3k^2-2k+\frac{k!}{(k-3)!}\\

k^3-3k^2+2k=\frac{k!}{(k-3)!}\\

(k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\~\\

LHS=x^3+3k^2+3k+1\\~\\

RHS=k+1+3k^2+3k+6*\frac{(k+1)!}{3!(k-2)!}\\

RHS=3k^2+4k+1+\frac{(k+1)!}{(k-2)!}\\

RHS=3k^2+4k+1+\frac{k!(k+1)}{(k-3)!(k-2)}\\

RHS=3k^2+4k+1+\frac{k+1}{k-2}*\frac{k!}{(k-3)!}\\

sub\\

RHS=3k^2+4k+1+\frac{k+1}{k-2}*(k^3-3k^2+2k)\\

RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k^2-3k+2)\\

RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k-2)(k-1)\\

RHS=3k^2+4k+1+(k+1)*k(k-1)\\

RHS=3k^2+4k+1+k^3-k\\

RHS=k^3+3k^2+3k+1\\

RHS=LHS

Melody Nov 2, 2020