((1 - 5/4)^3/(1 + 1/8 - 7/4) - 1 + 3/5)^2/((- 6/5)/(8/5))^3 + 1=
=((- 1/4)^3/(- 5/8) - 1 + 3/5)^2/(- 3/4)^3 + 1=
=((- 1/64)/(- 5/8) - 1 + 3/5)^2/(- 27/64) + 1=
=(1/40 - 1 + 3/5)^2/(- 27/64) + 1=
=(- 3/8)^2/(- 27/64) + 1=
=9/64/(- 27/64) + 1=
=- 1/3 + 1= 2/3
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$\small \left[\left(1-\dfrac{5}{4}\right)^3 : \left(1+\dfrac{1}{8}-\dfrac{7}{4}\right)-1+\dfrac{3}{5}\right]^2 : \left[\left(-\dfrac{6}{5}\right) : \left(\dfrac{8}{5}\right)\right]^3+1=$
$\small =\left[\left(\dfrac{4-5}{4}\right)^3 : \left(\dfrac{8+1-14}{8}\right)-1+\dfrac{3}{5}\right]^2 : \left[-\dfrac{\cancel6^3}{\cancel5_1}·\dfrac{\cancel5^1}{\cancel8_4}\right]^3+1=$
$\small =\left[\left(-\dfrac{1}{4}\right)^3 : -\dfrac{5}{8}-1+\dfrac{3}{5}\right]^2 : \left[-\dfrac{3}{1}·\dfrac{1}{4}\right]^3+1=$
$\small =\left[-\dfrac{1}{\cancel{64}_8}·-\dfrac{\cancel8^1}{5}-1+\dfrac{3}{5}\right]^2 : \left[-\dfrac{3}{4}\right]^3+1=$
$\small =\left[-\dfrac{1}{8}·-\dfrac{1}{5}-1+\dfrac{3}{5}\right]^2 : -\dfrac{27}{64}+1=$
$\small =\left[\dfrac{1}{40}-1+\dfrac{3}{5}\right]^2 ·-\dfrac{64}{27}+1=$
$\small =\left[\dfrac{1-40+24}{40}\right]^2 ·-\dfrac{64}{27}+1=$
$\small =\left[-\dfrac{\cancel{15}^3}{\cancel{40}_8}\right]^2 ·-\dfrac{64}{27}+1=$
$\small =\left[-\dfrac{3}{8}\right]^2 ·-\dfrac{64}{27}+1=$
$\small = \dfrac{\cancel9^1}{\cancel{64}_1} ·-\dfrac{\cancel{64}^1}{\cancel{27}_3}+1=$
$\small = \dfrac{1}{1} ·-\dfrac{1}{3}+1=$
$\small = -\dfrac{1}{3}+1=$
$\small = \dfrac{-1+3}{3}=$
$\small = \dfrac{2}{3}$