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I need help with this Trig Identity with steps please 1+cot^3x/1+cotx=cscx^2-cotx
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treat cot x as A, and it forms the form of B^3 + A^3 = ..., your B =1 , the same. just open and get the answer
$$\bf \cfrac{1+cot^3(x)}{1+cot(x)}=csc^2(x)-cot(x)\\ \text{as already suggested by Loser66}\\ a^3+b^3 = (a+b)(a^2-ab+b^2)\\ 1+cot^3(x) \implies 1^3 + cot^3(x)\\ \cfrac{1+cot^3(x)}{1+cot(x)} \implies \cfrac{\cancel{(1+cot(x))}(1^2-(1)cot(x) +cot^2(x)) }{\cancel{1+cot(x)}} $$ what does that leave you with?
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