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By Vieta's formulas, $a+b+c=\frac{3}{2},$ so $a+b-1=\left(\frac{3}{2}-c\right)-1=\frac{1}{2}-c.$ Writing similar equations for the other two terms, we get $$(a+b-1{)}^3+(b+c-1{)}^3+(c+a-1{)}^3={\left(\frac{1}{2}-a\right)}^3+{\left(\frac{1}{2}-b\right)}^3+{\left(\frac{1}{2}-c\right)}^3.$$Now, note that $\left(\frac{1}{2}-a\right)+\left(\frac{1}{2}-b\right)+\left(\frac{1}{2}-c\right)=\frac{3}{2}-(a+b+c)=0.$ It's a general fact that if $r+s+t=0,$ then $r^3+s^3+t^3=3rst$; this follows from the factorization identity $$r^3+s^3+t^3=3rst+(r+s+t)(r^2+s^2+t^2-rs-st-rt).$$Therefore, $${\left(\frac{1}{2}-a\right)}^3+{\left(\frac{1}{2}-b\right)}^3+{\left(\frac{1}{2}-c\right)}^3=3\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right).$$Finally, letting $p(x)=2x^3-3x^2+165x-4,$ we have $p(x)=2(x-a)(x-b)(x-c),$ so $$78=p(\frac{1}{2})=2\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right).$$Therefore the answer is $$3\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right)=\frac{3}{2}\cdot 78=\boxed{117}.$$