Question

Show that (a+b)^2 + (a-b)^2 = 2(a^2+b^2).

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This is my homework.​

Answer 1

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

Show that (a+b)^2 + (a-b)^2 = 2(a^2+b^2).

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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$\bold{ = > {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}$

$\bold{= > {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}$

$\bold{= > {(a + b)}^{2} + {(a - b)}^{2} = {a}^{2} + {b}^{2} + 2ab + {a}^{2} + {b}^{2} - 2ab}$

=>Here,2ab and -2ab gets cancelled.

$\bold{= {(a + b)}^{2} + {(a - b)}^{2} = 2 {a}^{2} + 2 {b}^{2} = 2( {a}^{2} + {b}^{2} )}$

=>Here,2 comes two times along with a and b so,take 2 as common.

$\bold{\red{∴ {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2})}}$

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