quicker way to do this?

[jigsai22]

(1001^2-999^2)/(101^2-99^2)

Anyone have a quicker way to solve this than just doing the calculation?

[kingrulz]

(1001^2-999^2)/(101^2-99^2)
= [(1000+1)^2 - (1000-1)^2] / [(100+1)^2 - (100-1)^2]

{use the identity: (a+b)^2 - (a-b)^2 = 4ab}

= 4*1000*1 / 4*100*1
= 10.

Cheers!

[Bell_Curve]

the fastest way is using (a^2-b^2) = (a+b)(a-b) identitiy

=(1001+999)(1001-999)/(101+99)(101-99)
=2000*2/200*2
=10

[jigsai22]

And now I am kicking myself. Thanks guys.