Exam recall of $\alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)+3\alpha\beta\gamma$

Question

From my book:

You can find a similar result for the sum of the cubes of a cubic equation by multiplying out $(\alpha + \beta + \gamma)^3.$

• The rules for sums of cubes: -
  • Quadratic: $\alpha^3 + \beta^3 = (\alpha + \beta)^3-3\alpha\beta(\alpha + \beta)$
  • Cubic: $\alpha^3 + \beta^3 + \gamma^3 = (\alpha+\beta+\gamma)^3 - 3(\alpha + \beta + \gamma)( \alpha\beta + \beta\gamma + \gamma\alpha) + 3\alpha\beta\gamma $

The quadratic one is easy to see, but I don't know how to see where the cubic one comes from in a simple way/ with as little working as possible? I tried a few ways of expanding $( \alpha + \beta + \gamma )^{3}$, but didn't get the result easily (or at all).

Answer 1

For an exam, probably memorising this cubic rule is best. But let's say you're in an exam and failed to memorise the cubic rule. Then you just need to do 2 things and then deriving the rule is easy.

1. Remember you can expand the brackets of $(\alpha + \beta + \gamma)^{3}$ as such:

$\quad(\alpha + \beta + \gamma)^{3} = \alpha^{3} + \beta^{3} + \gamma^{3} + 3( \alpha\beta^{2}+\alpha\gamma^{2}+\beta\alpha^{2}+\beta\gamma^{2}+\gamma\alpha^{2}+\gamma\beta^{2}) + 6\alpha\beta\gamma$,

which has 3 + 18 + 6 = 27 terms, so we haven't missed out any terms. And you can remember to use:

2. $\quad(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma) = 3\alpha\beta\gamma + (\alpha\beta^{2}+\alpha\gamma^{2}+\beta\alpha^{2}+\beta\gamma^{2}+\gamma\alpha^{2}+\gamma\beta^{2})$

Now eliminate $\alpha\beta^{2}+\alpha\gamma^{2}+\beta\alpha^{2}+\beta\gamma^{2}+\gamma\alpha^{2}+\gamma\beta^{2}$ from both equations.

Answer 2

Let $k=\beta+\gamma$. So, I have: $$(\alpha+\beta+\gamma)^3=(\alpha+k)^3=\alpha^3+k^3+3\alpha k^2+3\alpha^2k=\alpha^3+\beta^3+\gamma^3+3\gamma^2\beta+3\beta^2\gamma+3\alpha^2\gamma+3\alpha^2\beta+3\alpha\beta^2+3\alpha\gamma^2+6\alpha\beta\gamma=\alpha^3+\beta^3+\gamma^3+3\gamma(\gamma\beta+\alpha\beta+\gamma\alpha)+3\alpha(\alpha\gamma+\alpha\beta+\gamma\beta)+3\beta(\alpha\gamma+\alpha\beta+\gamma\beta)-3\alpha\beta\gamma=\alpha^3+\beta^3+\gamma^3+3(\alpha+\beta+\gamma)(\alpha\gamma+\alpha\beta+\gamma\beta)-3\alpha\beta\gamma$$

Answer 3

There is a multinomial formula. For three variables, it reads as $$(x+y+z)^n=\sum_{i+j+k=n}\frac{n!}{i!\,j!\,k!}\,x^i y^j z^k$$ and in particular \begin{multline}(x+y+z)^3=x^3+y^3+z^3\\+3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2+6xyz,\end{multline} which may be easier to remember as $$(x+y+z)^3=x^3+y^3+z^3+3\bigl(xy(x+y)+yz(y+z)+zx(z+x)\bigr)+6xyz,$$ and writing $x+y=(x+y+z)-z$, &c., we obtain $$(x+y+z)^3=x^3+y^3+z^3+3(xy+yz+zx)(x+y+z)-3xyz.$$

Answer 4

Just do it.

$(a + b +c)^3 = (a+b+c)(a+b+c)(a+b+c) =$

$a^3 + a^2b + a^2c + aba + ab^2 + abc + aca+acb + ac^2 + ba^2 + bab + bac +b^2a + b^3 +b^2c + bca + bcb + bc^2 + ca^2 + cab + cac + cba + cb^2 + cbc + c^2 a + c^2b+ c^3=$

$(a^3 + b^3 + c^3) + 3(a^2 b + a^2c + ab^2 +b^2c + ac^2+bc^2) +6abc=$

$(a^3 + b^3 + c^3) + 3([a^2b + a^2c + abc]+ [ab^2 + abc + b^2c] + [abc + ac^2 + bc^2] + 3abc =$

$(a^3 + b^3 + c^3) + 3(a(ab + ac +bc) + b(ab + ac +bc) + c(ab +ac +bc)) + 3abc =$

$(a^3 + b^3 +c^3) +3(a+b+c)(ab+ac + bc) + 3abc$

......

$(a+b+c)^3 = $ sum of all the cubes of all terms ($a^3 + b^3 +c^3)+$

$3$ ways of getting one of the terms times the square of the sum of the other two terms $

$3[ a(b+c)^2 + b(a+c)^2 + c(a+b)^2]$.

Meanwhile looking at each $a(b+c)(b+c)$ and $b(a+c)(a+c)$ and $c(a+b)(a+b)$ we have every term times the some of product of any two terms: That is to say, from $a(b+c)(b+c)$ we can extract $a(b^2)$ and $a(bc)$(twice) and $a(c^2)$ whereas for $b(a+c)(a+c)$ we can extract $a(ab)$ and $a(bc)$ twice and so on. We have three of each $a(b+c)^2$ so we can do this extraction for each $a,b,c$ so from $3[ a(b+c)^2 + b(a+c)^2 + c(a+b)^2]$ we can get $a(ab+bc+ac)$ and $b(ab+bc + ac)$ and $c(ab+bc+ac)$, as well as three extra $abc$s.

So this should intuitively give us the result.

==== third "what this finger painting is supposed to represent" explanation =====

$(a + b+ c)^3=(a+b+c)(a+b+c)(a+b+c)$ should be a sum of the $27$ different combinations of three terms. They are:

• terms of the form $x^3$. There are three of those; one for $x=a,b,c$
• terms of the form $x^2y$. for each $a,b,$ or $c$ there are two out of three of the triplets, $(a+b+c)$ where the two $x$s can come from, and the $y$ must come from the third triplet. That is to say. to get $\color{blue}{x^2y}$, we get it from $(\color{blue}x + \color{gray}y + \color{gray}z)(\color{blue}x + \color{gray}y + \color{gray}z)(\color{gray}x + \color{blue}y + \color{gray}z)$ and there are ${3\choose 2}={3\choose 1} = 3$ ways to choose which of $x$s and which $y$ to use. And of $a,b,c$ there are $3$ choices for the $x$ and $2$ remaining for the $y$. And of those $6$ there are $3$ choices of how you pick the $x$ and $y$ so those are $18$ of these $x^2y$s.
• the remaining six terms are the terms $abc$. It makes sense that there are $6! =6$ of these terms, because there are $3$ triplets to choose the $a$ from; once choosen there are $2$ remaining triplets to choose the $b$, etc.

The second item on the list can be written as $3(a^2(b+c) + b^2(a+c)+c^2(b+c)$ or purhaps more "balanced" as $3(a(ab+ac) + b(ab + bc) + c(ab + bc))$.

And we can "borrow " three of the $abc$s from the third item on the list so that $3(a(ab + ac + bc) + b(ab + ac + bc)+ c(ab + ac + bc) ) = 3(a+b+c)(ab + bc + ac)$

So ultimately

$(a+b+c)^3 = (a^3 + b^3 + c^3) + 3(a+b+c)(ab+bc + ac) + 3abc$.

That's nothing more or less than the sum of all combinations presented and factored in a fairly convenient and simple form.