$\frac{\sin^4x-\cos^4x}{\sin^2x+\sin x.\cos x}$
$25y^2+70yz+49^2$
$\int\:6\cos\:^3\left(x\right)dx$
$625\cdot9$
$\left(-9\right)-\left[-6-\left(3+2-1\right)+3\right]$
$1+\frac{-4x-3}{3+x^2}\:\left(\:8\right)$
$-1-\left(-6\right)+\left(+3\right)$
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