Question

Explain why each non-zero integer has two square roots but only one cube root.

Answer 1

if we have a number like say hmm 4, and we say hmmm √4 is ±2, it simply means, that if we multiply that number twice by itself, we get what's inside the root, we get the 4, so (+2)(+2) = 4, and (-2)(-2) = 4, recall that minus times minus = plus.

so, any when we're referring to even roots like $\bf \sqrt[2]{~~},\sqrt[4]{~~},\sqrt[6]{~~}....$, the positive number, that can multiply itself an even amount of times, will produce a valid value, BUT the negative number that multiply itself an even amount of times, will also produce a valid value.

now, that's is not true for odd roots like $\bf \sqrt[3]{~~},\sqrt[5]{~~},\sqrt[7]{~~}....$, because the multiplication of the negative number will not produce a valid value, let's put two examples on that.

$\bf \sqrt[3]{27}\implies \sqrt[3]{3^3}\implies 3\qquad because\qquad (3)(3)(3)=27 \\\\\\ however\qquad (-3)(-3)(-3)\ne 27~\hspace{8em}(-3)(-3)(-3)=-27 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt[3]{-125}\implies \sqrt[3]{-5^3}\implies -5\qquad because\qquad (-5)(-5)(-5)=-125 \\\\\\ however\qquad (5)(5)(5)\ne -125~\hspace{10em}(5)(5)(5)=125$

so, when the root is an odd root, you will always get only one number that will produce the radicand.