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