(a+b)=a²+b²+2ab derivation formulas
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(a+b)(a+b)=a^2+ab+ab+b^2
(a+b)^2=a^2+2ab+b^2
(a+b)^2=a^2+2ab+b^2
shivanshtripathi:
this is correct or wrong
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Solution ●●●●
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To prove that
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![(a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab](https://tex.z-dn.net/?f=%28a+%2B+b%29+%7B%7D%5E%7B2%7D+%3D+%7Ba%7D%5E%7B2%7D+%2B+%7Bb%7D%5E%7B2%7D+%2B+2ab)
PROOF :-
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![= (a + b)(a + b) \\ \\ = a(a + b) + b(a + b) \\ \\ = {a}^{2} + ab + ab + {b}^{2} \\ \\ = {a}^{2} + 2ab + {b}^{2} \\ \\ = {a}^{2} + {b}^{2} + 2ab = (a + b)(a + b) \\ \\ = a(a + b) + b(a + b) \\ \\ = {a}^{2} + ab + ab + {b}^{2} \\ \\ = {a}^{2} + 2ab + {b}^{2} \\ \\ = {a}^{2} + {b}^{2} + 2ab](https://tex.z-dn.net/?f=+%3D+%28a+%2B+b%29%28a+%2B+b%29+%5C%5C+%5C%5C+%3D+a%28a+%2B+b%29+%2B+b%28a+%2B+b%29+%5C%5C+%5C%5C+%3D+%7Ba%7D%5E%7B2%7D+%2B+ab+%2B+ab+%2B+%7Bb%7D%5E%7B2%7D+%5C%5C+%5C%5C+%3D+%7Ba%7D%5E%7B2%7D+%2B+2ab+%2B+%7Bb%7D%5E%7B2%7D+%5C%5C+%5C%5C+%3D+%7Ba%7D%5E%7B2%7D+%2B+%7Bb%7D%5E%7B2%7D+%2B+2ab)
I hope it will help you
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To prove that
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PROOF :-
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I hope it will help you
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