We know the following factorization:

**Proof 1:**

Let a, b, c be roots of a polynomial P(X) Then by fundamental theorem of algebra

.

Now, once again basic algebra says that as a, b, c are roots/solutions of the above:

=0$

Adding all the above:

So, we get

**Also, the above formula can be written as **

**Proof 2:**

Consider the following determinant D:

On adding all three columns to the first column: we know that the value of the determinant is unchanged: So we get the following:

. Note that columns 2 and 3 of the three by three determinant do not change.

On expanding the original determinant D, we get

Whereas we get from the other transformed but equal D:

So, that we again get

**Proof 3:**

Now, let us consider as a quadratic in a with b and c as parameters.

That is,

Then, the discriminant is given by

, which in turn equals:

Hence, the factorization of the above quadratic in a is given as:

So, the other non-trivial factorization of the above famous algebraic identity is:

where are cube roots of unity.

**Proof 4: **

Factorize the expression .

**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

is

**Proof 5:**

By Cramer’s rule, the eliminant is given by determinant .

On expansion using the first row:

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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