Aptitude - Surds and Indices - Discussion
Discussion Forum : Surds and Indices - General Questions (Q.No. 14)
14.
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xb |  |
(b + c - a) | . | |
xc | |
(c + a - b) | . | |
xa | |
(a + b - c) | = ? |
| xc | xa | xb |
Answer: Option
Explanation:
| Given Exp. |
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Discussion:
22 comments Page 1 of 3.
Kowsalya said:
5 months ago
@All.
(b - c) (b + c) - a (b - c) I don't understand this step. How are you multiplying (b-c) separately with a? Anyone explain it to me.
(b - c) (b + c) - a (b - c) I don't understand this step. How are you multiplying (b-c) separately with a? Anyone explain it to me.
Ajay said:
1 year ago
Very useful question and thanks for all the given explanation.
Raj said:
4 years ago
I can't get 3rd step, can anyone, explain to me? please.
(1)
Md. Zihadur Rahman said:
4 years ago
How can it be possible that (b - c) (b + c) - a (b-c) = (b-c) (b+c-a)?
Can you explain it, please?
Can you explain it, please?
(2)
Gautam said:
4 years ago
Here, (.) Means multiply right?
Jeevan Kumar said:
5 years ago
@All.
x(b2+bc-ab-cb-c2+ac) by following like this you will get finally x(0) = 1.
x(b2+bc-ab-cb-c2+ac) by following like this you will get finally x(0) = 1.
Amulya said:
5 years ago
Thanks all for explaining.
Sachin said:
7 years ago
I can't understand the 2nd step. Please, anyone explain it.
Chirag said:
7 years ago
@Rita.
It is actually [x^(b-c)]^(b+c-a).
According to law of INDICES, (a^m)^n= a^mn.
So, [x^(b-c)]^(b+c-a)=x^(b-c)(b+c-a).
It is actually [x^(b-c)]^(b+c-a).
According to law of INDICES, (a^m)^n= a^mn.
So, [x^(b-c)]^(b+c-a)=x^(b-c)(b+c-a).
Rita said:
7 years ago
I can't understand 2nd step, Please, someone explain to me.
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