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How do you prove $\frac{{tan}^{3} x - 1}{tan x - 1} = {tan}^{2} x + tan x + 1$ $(tan^3x - 1) /( tanx - 1) = tan^2x + tanx + 1$ ?

1 Answer

May 22, 2016

as directed below

Explanation:

Please factorize numerator following the formula
$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$ . you will get result.

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