## Various proofs of important algebraic identity: a^{3}+b^{3}+c^{3}-3abc

We know the following factorization: $a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Proof 1:

Let a, b, c be roots of a polynomial P(X) Then by fundamental theorem of algebra $P(X)=(X-a)(X-b)(X-c) = X^{3}-(a+b+c)X^{2}+(ab+bc+ca)X-abc$.

Now, once again basic algebra says that as a, b, c are roots/solutions of the above: $P(a)=a^{3}-(a+b+c)a^{2}+(ab+bc+ca)a-abc=0$ $P(b)=b^{3}-(a+b+c)b^{2}+(ab+bc+ca)b-abc=0$ $P(c)=c^{3}-(a+b+c)c^{2}+(ab+bc+ca)c-abc$=0\$ $0= a^{3}+b^{3}+c^{3}-(a+b+c)(a^{2}+b^{2}+c^{2})+(ab+bc+ca)(a+b+c)-3abc$

So, we get $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Also, the above formula can be written as $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(\frac{1}{2})((a-b)^{2}+(b-c)^{2}+(c-a)^{2})$

Proof 2:

Consider the following determinant D: $\left| \begin{array}{ccc} a & b & c\\c & a & b\\ b & c & a \end{array} \right|$

On adding all three columns to the first column: we know that the value of the determinant is unchanged: So we get the following: $D = \left| \begin{array}{ccc} a+b+c & b & c \\a+b+c & a & b\\a+b+c & c & a \end{array} \right|$. Note that columns 2 and 3 of the three by three determinant do not change.

On expanding the original determinant D, we get $D = a(a^{2}-bc)-b(ac-b^{2})+c(c^{2}-ab)$ $D= a^{3}-abc-bac+b^{3}+c^{3}-cab$ $D= a^{3}+b^{3}+c^{3}-3abc$

Whereas we get from the other transformed but equal D: $D =(a+b+c) \left| \begin{array}{ccc} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{array}\right|$ $D=(a+b+c)((a^{2}-bc)-b(a-b)+c(c-a))$

So, that we again get $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Proof 3:

Now, let us consider $E=a^{2}+b^{2}+c^{2}-ab-bc-ca$ as a quadratic in a with b and c as parameters.

That is, $E= a^{2} -(b+c)a + b^{2}+c^{2}-bc$

Then, the discriminant is given by $\triangle = (b+c)^{2}-4 \times 1 \times (b^{2}+c^{2}-bc)$, which in turn equals: $\triangle = (b^{2}+c^{2}+2bc)-4(b^{2}+c^{2}-bc) = -3b^{2}-3c^{2}+6bc=-3(b-c)^{2}$ $a_{1}, a_{2} = \frac{(b+c) \pm i\sqrt{3}(b-c)}{2}$ $a_{1}= b(\frac{1+i\sqrt{3}}{2})=c(\frac{1-i\sqrt{3}}{2}) = -b\omega - c\omega^{2}$ $a_{2}=b(\frac{1-i\sqrt{3}}{2})+c(\frac{1-i\sqrt{3}}{2})=-b\omega^{2}-c\omega$

Hence, the factorization of the above quadratic in a is given as: $a^{2}+b^{2}+c^{2}-ab-bc-ca = (a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)$

So, the other non-trivial factorization of the above famous algebraic identity is: $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+b\omega+ c\omega^{2})(a+b\omega^{2}+c\omega)$

where $1, \omega, \omega^{2}$ are cube roots of unity.

Proof 4:

Factorize the expression $a^{3}+b^{3}+c^{3}-3abc by (a+b+c)$.

Solution 4:

We can carry out the above polynomial division by considering the dividend to be a polynomial in a single variable, say a, (and assuming b and c are just parameters; so visualize them as arbitrary but fixed constants); further arrange the dividend in descending powers of a; so also arrange the divisor in descending powers of a (well, of course, it is just linear in a; and assume b and c are parameters also in dividend).

Proof 5:

Prove that the eliminant of $ax+cy+bz=0$ $cx+by+az=0$ $bx+ay+cz=0$

is $a^{3}+b^{3}+c^{3}-3abc=0$

Proof 5:

By Cramer’s rule, the eliminant is given by determinant $\left| \begin{array}{ccc}a & c & b \\ c & b & a\\b & a & c \end{array}\right|=0$.

On expansion using the first row: $a(bc-a^{2})-c(c^{2}-ab)+b(ac-b^{2})=0$ $a^{3}+b^{3}+c^{3}-3abc=0$ upon multiplying both the sides of the above equation by (-1). Of course, we have only been able to generate the basic algebraic expression but we have done so by encountering a system of three linear equations in x, y, z. (we could append any of the above factorization methods to this further!! :-)))

Cheers,

Nalin Pithwa

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