Utkrisht123 3 Report post Posted March 24, 2013 (edited) 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 March 24, 2013 by Utkrisht123 Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted March 24, 2013 Can't you prove it just by expanding the product? All the steps are reversible, so going either direction accomplishes the same thing. Share this post Link to post Share on other sites

0 James33 1 Report post Posted March 24, 2013 (a+b)(a-b)=a(a-b)+b(a-b)=a^2-ab+ba-b^2=a^2-b^2 Share this post Link to post Share on other sites

0 Prime 15 Report post Posted March 24, 2013 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) Share this post Link to post Share on other sites

0 vinay.singh84 0 Report post Posted March 25, 2013 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) Share this post Link to post Share on other sites

0 Perhaps check it again 22 Report post Posted January 26, 2014 (edited) 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 January 26, 2014 by Perhaps check it again Share this post Link to post Share on other sites

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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