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$\frac{{(x - 1)}^{2 n} {(x + 1)}^{n + 1}}{{(x - 1)}^{n} {(x + 1)}^{n}}$ $((x-1)^(2n)(x+1)^(n+1))/((x-1)^(n)(x+1)^n)$

1 Answer

Mar 13, 2018

${(x - 1)}^{n} (x + 1)$ $color(red)((x-1)^n(x+1)$

Explanation:

Remember that $\frac{a^{p}}{a^{q}} = a^{p - q}$ $(a^p)/(a^q)=a^(p-q)$

So

$\frac{{(x - 1)}^{2 n} {(x + 1)}^{n + 1}}{{(x - 1)}^{n} {(x + 1)}^{n}}$ $((x-1)^(2n)(x+1)^(n+1))/((x-1)^n(x+1)^n)$

$XXX = \frac{{(x - 1)}^{2 n}}{{(x - 1)}^{n}} \cdot \frac{{(x + 1)}^{n + 1}}{{(x + 1)}^{n}}$ $color(white)("XXX")=color(blue)(((x-1)^(2n))/((x-1)^n)) * color(magenta)(((x+1)^(n+1))/((x+1)^n))$

$XXX = {(x - 1)}^{n} \cdot (x + 1)$ $color(white)("XXX")=color(blue)((x-1)^n) * color(magenta)(""(x+1))$

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