**Problem 1:**

Prove that .

**Proof I:**

First carry out the following steps:

(a) Is the given expression symmetric?

(b) Is the given expression alternating?

(c) Is the given expression cyclic?

(d) Is the given expression homogeneous?

So here goes the proof:

Denote the expression by E. Clearly, the substitutions , give us . Hence, two of the factors are . Similarly, the substitution give us . So the other factor is .

The degree of the given expression E is 5. E is symmetric w.r.t. a and b; it is homogeneous. The factors we found above are . So where F(x,y) ought to be of degree 2, homogeneous and symmetric. So, let ; thus, we have

. Now, find k, A, and B by the undetermined coefficients.

Hence, you will get .

**Problem 2:**

Find the factors of .

**Solution 2:**

First carry out the following steps:

(a) Is the given expression symmetric w.r.t. a and b; b and c; c and a?

(b) Is the given expression alternating w.r.t. a and b; b and c; c and a?

(c) Is the given expression cyclic w.r.t. a, b and c?

(d) Is the given expression homogeneous and if so, what is the degree of each expression?

Denote the expression by E; then, E is a function of a which vanishes when , , , so the following are certainly factors of E: , and hence, E contains as a factor.

Now note that E is of fourth degree and symmetric, so the remaining factor is of the form : , where M is a coefficient to be determined.

So, we have now: , and you can easily find that , and hence,

More mathematical miscellany for IITJEE mathematics to follow!

Nalin Pithwa

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