$\tan\left(2x\right)=\frac{4}{\cot\left(x\right)\tan\left(x\right)}$
$a^{-2}x^{-1}\cdot\left(\frac{ax^{-2}}{x^{4}}\right)^{5}$
$\left(3x+4\right)^2-10x=9x^2+12x+16$
$55+-49+90$
$x^3+14x^2+13x$
$x^2-16+256$
$4x^2+12x-1$
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