Cyclic Fractions for IITJEE foundation maths

Consider the expression


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.


(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.


(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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