$\lim_{x\to\infty}\left(\frac{\left(8x-3\right)}{\left(8x+5\right)}\right)^{7x+1}$
$\left(a+b\right)^3\left(a+b\right)^3$
$\left(+6\right)\:\left(-3\right)\:\left(-9\right)\:\left(-1\right)\:=\:$
$\frac{\left(3x^4+3x^2+x-5\right)}{\left(x+2\right)}$
$3+2+5+3+2+7+1+2$
$0.7x-0.8y=1.6$
$8x^3+27b^6$
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