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:
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Discussion:
22 comments Page 1 of 3.
Mansi said:
1 decade ago
In the second step:
From the first term: x^(b-c)(b+c-a).
There are two brackets term, from the second bracket -a is taken out. As it comes out, it has to be multiplied by the first bracket.
So it gets multiplied and we get x^(b-c)(b+c)-a(b-c).
Similar thing is done for the rest.
From the first term: x^(b-c)(b+c-a).
There are two brackets term, from the second bracket -a is taken out. As it comes out, it has to be multiplied by the first bracket.
So it gets multiplied and we get x^(b-c)(b+c)-a(b-c).
Similar thing is done for the rest.
(1)
Ganesh said:
1 decade ago
Second step is the simple representation through indices and hence in the third step the exponents are multiplied and again with the help of indices the sum is solved it is a very simple but tricky question.
Hope you understand my explanation. All the best.
Hope you understand my explanation. All the best.
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.
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).
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)
Gaurav joshi said:
9 years ago
X^(b-c)((b+c)-a).
We get, X^(b-c)(b+c)-a(b-c).
I hope we all get understand by this.
We get, X^(b-c)(b+c)-a(b-c).
I hope we all get understand by this.
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.
Jegan said:
8 years ago
Next step they are taking (a+b)(b-C) = A*2-B*2 & adding in common X.
(1)
Anand kumar said:
8 years ago
I should not understand this.
Someone's can explain this properly?
Someone's can explain this properly?
Ammar said:
9 years ago
What does the dot (.) means between those numbers, help anybody?
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