How do you use the Binomial theorem to expand #(4-5i)^3#?

1 Answer
Binayaka C.
Jul 5, 2017

#(4-5i)^3 = -236-115i#

Explanation:

We know #(a+b)^n= nC_0 a^n*b^0 +nC_1 a^(n-1)*b^1 + nC_2 a^(n-2)*b^2+..........+nC_n a^(n-n)*b^n#

Here #a=4,b=-5i,n=3# We know, #nC_r = (n!)/(r!*(n-r)!#
#:.3C_0 =1 , 3C_1 =3, 3C_2 =3,3C_3 =1 ; i^2=-1 ,i ^3 = -i #

#:.(4-5i)^3 = 4^3+3*4^2*(-5i) +3*4*(-5i)^2+(-5i)^3# or

#(4-5i)^3 = 64-240i+300i^2-125i^3# or

#(4-5i)^3 = 64-240i-300+125i# or

#(4-5i)^3 = -236-115i# [Ans]

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