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Aptitude - Surds and Indices - Discussion

Discussion Forum : Surds and Indices - General Questions (Q.No. 14)
Surds and Indices
14.
xb (b + c - a) . xc (c + a - b) . xa (a + b - c) = ?
xc xa xb
xabc
1
xab + bc + ca
xa + b + c
Answer: Option
Explanation:
Given Exp.
= x(b - c)(b + c - a) . x(c - a)(c + a - b) . x(a - b)(a + b - c)
= x(b - c)(b + c) - a(b - c)  .  x(c - a)(c + a) - b(c - a)
   .  x(a - b)(a + b) - c(a - b)
= x(b2 - c2 + c2 - a2 + a2 - b2)  .   x-a(b - c) - b(c - a) - c(a - b)
= (x0 x x0)
= (1 x 1) = 1.
Discussion:
22 comments Page 3 of 3.
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.
Dhanashree said:   1 decade ago
Little complicated sum please explain from 2nd step.
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