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a) Given $x^\\var{intpower}=\\var{intrhs}$, we can take the $\\var{intpower}$nd rd th  root of both sides.

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 $x^\\var{intpower}$ $=$ $\\var{intrhs}$ $\\sqrt[\\var{intpower}]{x^\\var{intpower}}$ $=$ $\\sqrt[\\var{intpower}]{\\var{intrhs}}$ $x$ $=$ $\\var{intsoln}$
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b) Given $\\simplify{{bxcoeff}y^{bpower}+{bb}}=\\var{bc}$, we can rearrange the equation to get $y^\\var{bpower}$ by itself and then we can take the $\\var{bpower}$nd rd th  root of both sides to get $y$ by itself.

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 $\\simplify{{bxcoeff}y^{bpower}+{bb}}$ $=$ $\\var{bc}$ $\\simplify{{bxcoeff}y^{bpower}}$ $=$ $\\simplify[basic]{{bc}-{bb}}$ $\\simplify{{bxcoeff}y^{bpower}}$ $=$ $\\simplify{{bc-bb}}$ $y^\\var{bpower}$ $=$ $\\simplify[!basic]{{bc-bb}/{bxcoeff}}$ $y^\\var{bpower}$ $=$ $\\simplify{{bc-bb}/{bxcoeff}}$ $\\sqrt[\\var{bpower}]{y^\\var{bpower}}$ $=$ $\\sqrt[\\var{bpower}]{\\var{brhs}}$ $y$ $=$ $\\var{bsoln}$
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c) Given $\\simplify{{cxcoeff}z^{cpower}+{cb}}=\\var{cc}$, we can rearrange the equation to get $z^\\var{cpower}$ by itself and then we can take the $\\var{cpower}$nd rd th  root of both sides to get $z$ by itself.

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 $\\simplify{{cxcoeff}z^{cpower}+{cb}}$ $=$ $\\var{cc}$ $\\simplify{{cxcoeff}z^{cpower}}$ $=$ $\\simplify[basic]{{cc}-{cb}}$ $\\simplify{{cxcoeff}z^{cpower}}$ $=$ $\\simplify{{cc-cb}}$ $z^\\var{cpower}$ $=$ $\\simplify[!basic]{{cc-cb}/{cxcoeff}}$ $z^\\var{cpower}$ $=$ $\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}$ $\\sqrt[\\var{cpower}]{z^\\var{cpower}}$ $=$ $\\sqrt[\\var{cpower}]{\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}}$ $z$ $=$ $\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}^{1/{cpower}}}$
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d) Given $\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}=\\var{dc}$, we can rearrange the equation to get $\\simplify{(z+{db})^{bpower}}$ by itself, then we can take the $\\var{bpower}$nd rd th  root of both sides to get $\\simplify{z+{db}}$ by itself, and then rearrange to get $z$ by itself.

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 $\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}$ $=$ $\\var{dc}$ $\\simplify{(z+{db})^{bpower}}$ $=$ $\\simplify[basic]{{dc}*{ddenom}}$ $\\simplify{(z+{db})^{bpower}}$ $=$ $\\var{dc*ddenom}$ $\\sqrt[\\var{bpower}]{\\simplify{(z+{db})^{bpower}}}$ $=$ $\\sqrt[\\var{bpower}]{\\var{dc*ddenom}}$ $\\simplify{z+{db}}$ $=$ $\\sqrt[\\var{bpower}]{\\var{dc*ddenom}}$ $z$ $=$ $\\simplify{root({dc*ddenom},{bpower})-{db}}$ $z$ $=$ $\\simplify{{dc*ddenom}^(1/{bpower})-{db}}$
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If  $x^\\var{intpower}=\\var{intrhs}$, then $x=$ [[0]].

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If  $\\simplify{{bxcoeff}y^{bpower}+{bb}}=\\var{bc}$, then $y=$ [[0]].

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For this question input the exact value by using a fractional power to indicate a root. For example, if the answer was $\\sqrt[3]{\\frac{35}{11}}$, then enter  (35/11)^(1/3).

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If  $\\simplify{{cxcoeff}z^{cpower}+{cb}}=\\var{cc}$, then $z=$ [[0]].

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For this question input the exact value by using a fractional power to indicate a root. For example, if the answer was $\\sqrt[3]{\\frac{35}{11}}$, then enter  (35/11)^(1/3).

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If  $\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}=\\var{dc}$, then $z=$ [[0]].

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Questions to test if the student knows the inverse of an odd power (and how to solve equations that contain a single power that is odd).

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