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Exulus · 23-08-2003, 22:37

mmmm, thats confusing but in this last step:

2(a^2 - ab) = 1(a^2 - ab)

You say that a and b are the same, and so the bits in the brakets will always equal 0, so you are just saying 0=0?

23-08-2003, 22:37	#25
Exulus Inactive Join Date: Jun 2003 Location: South Age: 40 Posts: 214	mmmm, thats confusing but in this last step: 2(a^2 - ab) = 1(a^2 - ab) You say that a and b are the same, and so the bits in the brakets will always equal 0, so you are just saying 0=0?