Posted 24 Mar 2013 (edited) · Report post All of you know that a^{2 }- b^{2} = (a+b)(a-b) I know only two method to prove it. a^{2 }- b^{2 }a^{2 }- b^{2} a^{2 }- b^{2 }+ ab - ab a^{2 }- b^{2} + 2ab -2ab -2b^{2} a^{2 }+ ab - ab - b^{2 }(a+b)^{2} - 2b(a+b) a(a+b) - b(a+b) (a+b)(a+b) - 2b(a+b) (a-b)(a+b) (a+b)(a+b-2b) (a+b)(a-b) Apart form these two method are there any other method to prove it. Edited 24 Mar 2013 by Utkrisht123 0 Share this post Link to post Share on other sites

0 Posted 24 Mar 2013 · Report post Can't you prove it just by expanding the product? All the steps are reversible, so going either direction accomplishes the same thing. 0 Share this post Link to post Share on other sites

0 Posted 24 Mar 2013 · Report post (a+b)(a-b)=a(a-b)+b(a-b)=a^2-ab+ba-b^2=a^2-b^2 0 Share this post Link to post Share on other sites

0 Posted 24 Mar 2013 · Report post Perhaps, OP is looking for this: Suppose, a = b + k, then k = a - b. a^{2} - b^{2} = (b+k)^{2} - b^{2} = b^{2} + 2bk + k^{2} - b^{2}= k(2b+k) = k(b+k+b)=(a-b)(a+b) 0 Share this post Link to post Share on other sites

0 Posted 25 Mar 2013 · Report post a^{2} - b^{2} (a + b - b)^{2} - b^{2} (a + b)^{2 }+ b^{2} - 2b(a + b) - b^{2} (a + b)^{2} - 2b(a + b) (a + b)(a + b -2b) (a + b)(a - b) 0 Share this post Link to post Share on other sites

0 Posted 26 Jan 2014 (edited) · Report post In post # 1 in the right-hand column under "a^2 - b^2," the next line should be "a^2 + b^2 + 2ab - 2ab - 2b^2." Edited 26 Jan 2014 by Perhaps check it again 0 Share this post Link to post Share on other sites

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All of you know that a

^{2 }- b^{2}= (a+b)(a-b)I know only two method to prove it.

a

^{2 }- b^{2 }a^{2 }- b^{2}a

^{2 }- b^{2 }+ ab - ab a^{2 }- b^{2}+ 2ab -2ab -2b^{2}a

^{2 }+ ab - ab - b^{2 }(a+b)^{2}- 2b(a+b)a(a+b) - b(a+b) (a+b)(a+b) - 2b(a+b)

(a-b)(a+b) (a+b)(a+b-2b)

(a+b)(a-b)

Apart form these two method are there any other method to prove it.

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