How do you evaluate #(i\sqrt { 3} ) ^ { 3} - 9( i\sqrt { 3} )#?

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Answer 1 · 2017-09-14T07:29:49

#=12isqrt3#

#(isqrt3)^3-9(isqrt3)#

#(1)#factorise by common factors

#=(isqrt3)[(isqrt3)^2)-9]#

#(2)#Now assuming #i=sqrt(-1)#

we have

#(isqrt3)(-3-9)=isqrt3xx(-12)#

#=-12isqrt3#