# Aptitude - Surds and Indices - Discussion

Discussion Forum : Surds and Indices - General Questions (Q.No. 8)
8.
 1 + 1 + 1 = ? 1 + x(b - a) + x(c - a) 1 + x(a - b) + x(c - b) 1 + x(b - c) + x(a - c)
0
1
xa - b - c
None of these
Explanation:

Given Exp. =
1  +  1  +  1
 1 + xb + xc xa xa
 1 + xa + xc xb xb
 1 + xb + xa xc xc

 = xa + xb + xc (xa + xb + xc) (xa + xb + xc) (xa + xb + xc)

 = (xa + xb + xc) (xa + xb + xc)

= 1.

Discussion:
24 comments Page 1 of 3.

I didn't understand the second line.

@raju

We took the lcm of the denominator of the expression 1+x^b/x^a...

Have any short cut method?

It is the shortcut and easiest way to solve this problem.

I did not understood second step from last.

Can you please explain in detail(only last two steps)

How do you consider x^a+x^b+x^c in numerator ?(x^a+x^b+x^c/x^a+x^b+x^c)

The second step is derived from first step by multiplying numerator & denominator of each of the 3 components with x^a, x^b & x^c respectively.

How do you solve the second step? Really confusing, please do it the long way so we can understand.

Abhijeet singh said:   1 decade ago
In the second line just take the LCM. and make the adjustment accordingly.
Its very simple try it on paper itself.
You guys will get the answer.

Thank you.

Shekhar kushwaha said:   9 years ago
What if x = 0?

KazirangaUniv said:   9 years ago
From the second step:

x^a(x^a+x^b+x^c)^2+x^b(x^a+x^b+x^c)^2+x^c(x^a+x^b+x^c)^2/(x^a+x^b+x^c)^3

Then take (x^a+X^b+x^c)^2 common from numerator and denominator and cut them both from top and bottom. Then we get step 3.