$$9+\{6+7aF3-(9+5-(+21\}$$
$45$
$$27+6+7\times 3-\left(9+2-\frac{6}{3}\right)$$
$$27+6+21-\left(9+2-\frac{6}{3}\right)$$
$$27+27-\left(9+2-\frac{6}{3}\right)$$
$$27+27-\left(11-\frac{6}{3}\right)$$
$$27+27-\left(11-2\right)$$
$$27+27-9$$
$$27+18$$
$$45$$
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$3^{2}\times 5$
