Rewriting using standard symbols:
#color(white)("XXXX")##(-105 div 35 + 6)(9)#
Everything inside the parentheses is done first with parentheses evaluated left to right
So #(-105 div 35 + 6)# is evaluated first
#color(white)("XXXX")#Within the parentheses multiplication and division is done before addition and subtraction
#color(white)("XXXX")#So #-105 div 35# is done first
#color(white)("XXXX")##color(white)("XXXX")##-105 div 35 = -3#
#color(white)("XXXX")##(-105div 35 + 6)# therefore becomes #(-3 +6)#
#color(white)("XXXX")##color(white)("XXXX")##-3+6 = 3#
#color(white)("XXXX")##(-105 div 35 +6)# becomes #(-3+6)# becomes #(3)# or just #3#
Moving to the second set of parentheses:
we have nothing to evaluate except #(9) = 9#
We are left with an implied multiplication
#color(white)("XXXX")##3xx9 = 27#
