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Question

Factorise:
x3+125
  1. (x+5)(x25x+25)
  2. (x+3)(x25x+25)
  3. (x+5)(x25x+25)
  4. (x+5)(x210x+25)


Solution

The correct option is A (x+5)(x25x+25)
We have x3+125,
125=5×5×5
This can be written as x3+53
Comparing x3+53 with a3+b3
we get a = x, b = 5 
Using identity,
a3+b3=(a+b)(a2ab+b2)
x3+53=(x+5)(x2(x)(5)+52)
x3+53=(x+5)(x25x+25)

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