Question

# ${\text{a}}^{2}{\text{+b}}^{2}\text{isequalto?}$

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Solution

## To solve this we use some identities in some steps :Step 1: Here we use $\left({\text{a+b)}}^{2}{\text{=a}}^{2}{\text{+b}}^{2}\text{+2ab}$ ${\text{a}}^{2}{\text{+b}}^{2}{\text{=(a+b)}}^{2}\text{-2ab}$(after shifting the sides)Step 2: Now we use the identity $\left({\text{a-b)}}^{2}{\text{=a}}^{2}{\text{+b}}^{2}\text{-2ab}$ ${\text{a}}^{2}{\text{+b}}^{2}{\text{=(a-b)}}^{2}\text{+2ab}$Step 3: Now we add the equation obtained in these two steps ${\text{a}}^{2}{\text{+b}}^{2}{\text{+a}}^{2}{\text{+b}}^{2}{\text{=(a+b)}}^{2}{\text{-2ab+(a-b)}}^{2}\text{+2ab}\phantom{\rule{0ex}{0ex}}2{\text{a}}^{2}{\text{+2b}}^{2}{\text{=(a+b)}}^{2}{\text{+(a-b)}}^{2}\phantom{\rule{0ex}{0ex}}{\text{2(a}}^{2}{\text{+b}}^{2}{\text{)=(a+b)}}^{2}{\text{+(a-b)}}^{2}\phantom{\rule{0ex}{0ex}}{\text{(a}}^{2}{\text{+b}}^{2}\text{)=}\frac{{\text{(a+b)}}^{2}{\text{+(a-b)}}^{2}}{2}$Hence, we get $\left({\text{a}}^{2}{\text{+b}}^{2}\text{)=}\frac{{\text{(a+b)}}^{2}{\text{+(a-b)}}^{2}}{2}$.

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