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)} 
Answer: Option
Explanation:
Given Exp. = 

=  x^{a}  +  x^{b}  +  x^{c} 
(x^{a} + x^{b} + x^{c})  (x^{a} + x^{b} + x^{c})  (x^{a} + x^{b} + x^{c}) 
=  (x^{a} + x^{b} + x^{c}) 
(x^{a} + x^{b} + x^{c}) 
= 1.
Discussion:
24 comments Page 1 of 3.
Ekta said:
7 years ago
In the second line,the denominator is solved as:
(1+x^b/x^a+x^c/x^a) (solving 1st denominator of the first fraction)
Which comes as( x^a+x^b+x^c)/x^a.
Now, the first fraction is : 1/((x^a+x^b+x^c)/x^a) which can be also written as x^a/(x^a+x^b+x^c).
Similarly other two fractions are also solved and finally x^a/(x^a+x^b+x^c) + x^b/(x^a+x^b+x^c) + x^c/(x^a+x^b+x^c) = 1.
(1+x^b/x^a+x^c/x^a) (solving 1st denominator of the first fraction)
Which comes as( x^a+x^b+x^c)/x^a.
Now, the first fraction is : 1/((x^a+x^b+x^c)/x^a) which can be also written as x^a/(x^a+x^b+x^c).
Similarly other two fractions are also solved and finally x^a/(x^a+x^b+x^c) + x^b/(x^a+x^b+x^c) + x^c/(x^a+x^b+x^c) = 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).
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).
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.
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.
Arshad said:
1 decade ago
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)
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)
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.
Its very simple try it on paper itself.
You guys will get the answer.
Thank you.
Sagar said:
1 decade ago
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.
Anagha said:
1 decade ago
How do you solve the second step? Really confusing, please do it the long way so we can understand.
Achitsa said:
8 years ago
First solve the denominator using LCM the answer you get. Find its reciprocal.
Kunaman said:
1 decade ago
@raju
We took the lcm of the denominator of the expression 1+x^b/x^a...
We took the lcm of the denominator of the expression 1+x^b/x^a...
Garima said:
7 years ago
Short cut method:
Let a=b=c,
We get , 1/3 +1/3 +1/3 = 3/3 =1.
Let a=b=c,
We get , 1/3 +1/3 +1/3 = 3/3 =1.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers