Solution 2.1:2e

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We expand the two quadratics using the squaring rule, and then sum the result
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We expand the two quadratics using the squaring rule, and then sum the result.
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<math> \qquad \begin{align}(a+b)^2+(a-b)^2 &= (a^2+2ab+b^2)+(a^2-2ab+b^2)\\
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{{Displayed math||<math>\begin{align}
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(a+b)^2+(a-b)^2 &= (a^2+2ab+b^2)+(a^2-2ab+b^2)\\
&= a^2+2ab+b^2+a^2-2ab+b^2 \\
&= a^2+2ab+b^2+a^2-2ab+b^2 \\
&= a^2+a^2+2ab-2ab+b^2+b^2\\
&= a^2+a^2+2ab-2ab+b^2+b^2\\
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&= 2a^2 +2b^2
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&= 2a^2 +2b^2\,\textrm{.}
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\end{align}
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\end{align}</math>}}
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</math>
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<!--<center> [[Image:2_1_2e.gif]] </center>-->
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Current revision

We expand the two quadratics using the squaring rule, and then sum the result

\displaystyle \begin{align}

(a+b)^2+(a-b)^2 &= (a^2+2ab+b^2)+(a^2-2ab+b^2)\\ &= a^2+2ab+b^2+a^2-2ab+b^2 \\ &= a^2+a^2+2ab-2ab+b^2+b^2\\ &= 2a^2 +2b^2\,\textrm{.} \end{align}

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