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Решите задачу:

$\tt \bigg(\cfrac{79^3-41^3}{38}+79\cdot41\bigg):(133.5^2-58.5^2) =\\\\\\= \bigg(\cfrac{(79-41)(79^2+79\cdot41+41^2)}{38}+79\cdot41\bigg):(133.5^2-58.5^2) =\\\\\\ = \bigg(\cfrac{38(79^2+79\cdot41+41^2)}{38}+79\cdot41\bigg):(133.5-58.5)(133.5+58.5)=\\\\\\=(79^2+79\cdot41+41^2+79\cdot41):(75\cdot192)=\\\\\\=\cfrac{(79+41)^2}{75\cdot192}= \cfrac{120^2}{75\cdot192}=\cfrac{(2^3\cdot3\cdot5)^2}{5^2\cdot3\cdot2^6\cdot3}=\cfrac{2^6\cdot3^2\cdot5^2}{5^2\cdot3^2\cdot2^6}=1$