## Algebra 1: Common Core (15th Edition)

$a.$ Conjecture: (the cube root of a number) = ($\displaystyle \frac{1}{3}$ power of a number) $b.$ Our conjecture works on $8$ and $27.$
$a.$ Generalizing on the pattern we observed in exercises 1-5: (the square root of a number) = ($\displaystyle \frac{1}{2}$ power of a number), We hypothesize that: (the cube root of a number) = ($\displaystyle \frac{1}{3}$ power of a number) $b.$ We know that $2^{3}=8$, so $\sqrt[3]{8}=2$ The calculator returns $8^{\wedge}(1\div 3)=2$ as well. We know that $3^{3}=27$, so $\sqrt[3]{27}=3$ The calculator returns $27^{\wedge}(1\div 3)=3$ as well So, our conjecture works for $8$ and $27.$