# 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.

Favour said:   8 years ago
I do not understand any of them.

Rutuja said:   2 years ago
It's great. Thanks all.

John said:   2 years ago
Thanks all for explaining the answer.

Shantanu Gupta said:   4 years ago
Here,

a, b, and c in cyclic order = 1.

SenthilRaja said:   5 years ago
Step 1: (1/(1+(x^b/x^a)+(x^c+x^a)))+(1/(1+(x^a/x^b)+(x^c+x^b)))+(1/(1+(x^b/x^c)+(x^a+x^c)))

Step 2: (1/((x^a+x^b+x^c)/x^a))+(1/((x^b+x^a+x^c)/x^b))+(1/((x^c+x^b+x^a)/x^c))

Step 3: (x^a/(x^a+x^b+x^c)) +(x^b/(x^a+x^b+x^c))+(x^c/(x^a+x^b+x^c)) Note: Take LCM in Denominator which is in (step 2 )

Step 4: (x^a+x^b+x^c)/(x^a+x^b+x^c) = 1 (Answer).

Ragav said:   6 years ago
Here, assign x=1.

Aarambh said:   6 years ago
I cannot understand the second step. Please explain me.

Ayush said:   6 years ago
@Garima

Garima said:   7 years ago
Short cut method.

Let a=b=c=1.
We get 1/3 +1/3 +1/3 =3/ 3 =1.

Garima said:   7 years ago
Short cut method:
Let a=b=c,
We get , 1/3 +1/3 +1/3 = 3/3 =1.