# Exercise ID134

Algebra → Exponents → Powers
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Evaluate the following expression:

${\left(\frac{{2}^{x}+{2}^{-x}}{2}\right)}^{2}-{\left(\frac{{2}^{x}-{2}^{-x}}{2}\right)}^{2}$

${\left(\frac{{2}^{x}+{{2}}^{-x}}{2}\right)}^{2}-{\left(\frac{{2}^{x}-{{2}}^{-x}}{2}\right)}^{2}$ $=\frac{{\left({2}^{x}+\frac{1}{{2}^{x}}\right)}^{2}}{{2}^{2}}-\frac{{\left({2}^{x}-\frac{1}{{2}^{x}}\right)}^{2}}{{2}^{2}}$ $=\frac{{\left(\frac{{2}^{x}·{2}^{x}+1}{{2}^{x}}\right)}^{2}}{\frac{{2}^{2}}{1}}-\frac{{\left(\frac{{2}^{x}·{2}^{x}-1}{{2}^{x}}\right)}^{2}}{\frac{{2}^{2}}{1}}$ $=\frac{{\left({2}^{2x}+1\right)}^{2}}{{2}^{2}·{2}^{2x}}-\frac{{\left({2}^{2x}-1\right)}^{2}}{{2}^{2}·{2}^{2x}}$ $=\frac{{\left({2}^{2x}+1\right)}^{2}-{\left({2}^{2x}-1\right)}^{2}}{{2}^{2x+2}}$ $=\frac{{\left({2}^{2x}\right)}^{2}+2·{2}^{2x}·1+1-\left[{\left({2}^{2x}\right)}^{2}-{2}^{2x}·2+1\right]}{{2}^{·+2}}$ $=\frac{{{2}}^{4x}+{2}^{2x+1}{+}{1}{-}{{2}}^{4x}+{2}^{2x+1}{-}{1}}{{2}^{2x-2}}$ $=\frac{{2}{·}{{2}}^{2x+1}}{{2}^{2x+2}}$ $=\frac{{{2}}^{2x+1+1}}{{2}^{2x+2}}$ $=\frac{{2}^{2x+2}}{{2}^{2x+2}}$ $=1$
${\left(\frac{{2}^{x}+{2}^{-x}}{2}\right)}^{2}-{\left(\frac{{2}^{x}-{2}^{-x}}{2}\right)}^{2}=1$

${a}^{-n}=\frac{1}{{a}^{n}};{\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^{n}$

${a}^{n}·{a}^{m}={a}^{n+m}$