Question

I need help verifying
$\frac{1 + { \cot(x) }^{3} }{1 + \cot(x) } + \cot(x) = 1 + { \cot(x) }^{2}$

Answer 1

$\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad (a+b)(a^2-ab+b^2)= a^3+b^3 \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad (a-b)(a^2+ab+b^2)= a^3-b^3\\\\ -------------------------------\\\\ \cfrac{1+cot^3(x)}{1+cot(x)}+cot(x)=1+cot^2(x) \\\\\\ \cfrac{1^3+cot^3(x)}{1+cot(x)}+cot(x)=1+cot^2(x)\\\\ -------------------------------\\\\ \cfrac{\underline{[1+cot(x)]}~~[1^2-[1cot(x)]+cot^2(x)]}{\underline{1+cot(x)}}+cot(x) \\\\\\ 1-cot(x)+cot^2(x)+cot(x)\implies 1+cot^2(x)$