## Cyclic Fractions for IITJEE foundation maths

Consider the expression

$\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}$

Here, in finding the LCM of the denominators, it must be observed that there are not six different compound factors to be considered; for, three of them differ from the other three only in sign.

Thus,

$(a-c) = -(c-a)$

$(b-a) = -(a-b)$

$(c-b) = -(b-c)$

Hence, replacing the second factor in each denominator by its equivalent, we may write the expression in the form

$-\frac{1}{(a-b)(c-b)}-\frac{1}{(b-c)(a-b)}-\frac{1}{(c-a)(b-c)}$ call this expression 1

Now, the LCM is $(b-c)(c-a)(a-b)$

and the expression is $\frac{-(b-c)-(c-a)-(a-b)}{(b-c)(c-a)(a-b)}=0$.,

Some Remarks:

There is a peculiarity in the arrangement of this example, which is desirable to notice. In the expression 1, the letters occur in what is known as cyclic order; that is, b follows a, a follows c, c follows b. Thus, if a, b, c are arranged round the circumference of a circle, if we may start from any letter and move round in the direction of  the arrows, the other letters follow in cyclic  order; namely, abc, bca, cab.

The observance of this principle is especially important in a large class of examples in which the differences of three letters are involved. Thus, we are observing cyclic order when we write $b-c$, $c-a$, $a-b$, whereas we are violating order by the use of arrangements such as $b-c$, $a-c$, $a-b$, etc. It will always be found that the work is rendered shorter and easier by following cyclic order from the beginning, and adhering to it throughout the question.

Homework:

(1) Find the value of $\frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}$

2) Find the value of $\frac{b}{(a-b)(a-c)} + \frac{c}{(b-c)(b-a)} + \frac{a}{(c-a)(c-b)}$

3) Find the value of $\frac{z}{(x-y)(x-z)} + \frac{x}{(y-z)(y-x)} + \frac{y}{(z-x)(z-y)}$

4) Find the value of $\frac{y+z}{(x-y)(x-z)} + \frac{z+x}{(y-z)(y-x)} + \frac{x+y}{(z-x)(z-y)}$

5) Find the value of $\frac{b-c}{(a-b)(a-c)} + \frac{c-a}{(b-c)(b-a)} + \frac{a-b}{(c-a)(c-b)}$

More later,

Nalin Pithwa

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