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

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Answer 1 · 2017-07-05T12:38:29

${(4 - 5 i)}^{3} = - 236 - 115 i$ $(4-5i)^3 = -236-115i$

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]

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