How do you evaluate #\frac { 1} { 2^ { 3} } ( - 32+ 8) - ( - 3) ^ { 3} \div 9#?

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Answer 1 · 2017-03-22T13:12:48

#=0#

#1/2^3(-32+8)-(-3^3)-:9#

#:.=1/2^3(-24)-(-27)-:9)#

#:.=1/2^3(-2^3 xx 3^1)-(-3)#

#:.=1/cancel(2^3)^color(red)1 xx -cancel(2^3)^color(red)1/1 xx 3^1/1+3#

#:.=1 xx -1 xx 3+3#

#:.=-3+3#

#:.=0#