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Question

Factorise a67a38.
 
  1. (a2)(a2+4)(a+1)(a3a1)
  2. (a2)(a2+a+2)(a+1)(a2a1)
  3. (a+2)(a2+a+4)(a+1)(a2+a+1)
  4. (a2)(a2+2a+4)(a+1)(a2a+1)


Solution

The correct option is D (a2)(a2+2a+4)(a+1)(a2a+1)
a67a38
=(a3)27a38
Let a3=x
Now,
x27x8
=x28x+x8
=x(x8)+1(x8)
=(x8)(x+1)
=(a38)(a3+1)
=(a323)(a3+1)
We know that,
{x3y3=(xy)(x2+xy+y2) }
{x3+y3=(x+y)(x2xy+y2) }
=(a2)(a2+2a+22)(a+1)(a2a+1)
=(a2)(a2+2a+4)(a+1)(a2a+1)

 

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