$2\sin^2\left(\frac{x}{4}\right)$
$\lim_{x\to\infty}\left(\frac{x^2+5}{x+6}\right)$
$\left(3u^4v^5\right)\left(7u^2v^3\right)$
$\left(1-n^4\right)\left(1-n^2\right)$
$4+2t^3$
$\left(2cosx+3sinx\right)^2=13$
$6x^2+16+8$
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