How do you simplify #12sqrt(-2)^12#?

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Answer 1 · 2016-03-21T13:11:31

#12sqrt((-2)^12)=3*2^8=768#

or

#12sqrt(-2^12)=768i in CC#

in #RR# the expression you writed has no semplification, but if you intend:

#12sqrt(-2^12)=12sqrt((-2)^12)#

You can rewrite expression as:

#12 sqrt((-1)^12*2^12)=12sqrt(2^12)=12sqrt((2^6)^2)=12*2^6=3*2^2*2^6=3*2^8=768#

Indeed, that's different from

#12sqrt(-2^12)=12sqrt(-1)*sqrt(2^12)=12*i*sqrt((2^6)^2)=768i#

with #i=sqrt(-1)# as immaginary unit of #CC#