Solve, please: \(\displaystyle{\left({u}-{2}\right)}^{{3}}+{40}={0}\), where u is a

Jackson Floyd

Jackson Floyd

Answered question

2022-03-21

Solve, please: (u2)3+40=0, where u is a real number

Answer & Explanation

clarkchica44klt

clarkchica44klt

Beginner2022-03-22Added 17 answers

(u2)3+40=0
Use odd root properl. If n is odd positive integer, then for any real number k xn=k is equivalent to x=kn
Thus, put foirst u2=x
And we get x=(40)1/3=403
Convert given number to polar form
40=40isin(0)+40cos(0)
All n-th roots of complex number:
γ(cosαisinα) are given by
γn(cos(α+2πkn)+isin(α+2πkn))
k=0,1,2,3,,n1,n,
We know tthat r=40,α=0,n=3
Thus, 
1)k=0
403(cos(0+2π(0)3)+isin(0+2π(0)3))
=253(cos0+isin0)=253
2)k=1
403(cos(0+2π(1)3)+isin(0+2π(1)3))
=253(cos(2π3)+isin(2π3))
=-53+25i3
3)k=2
403(cos(0+2π(2)3)+isin(0+2π(2)3))
=253(cos(4π3)+isin(4π3))
=-53-35i3
Thus, roots: x1=253
x2=-53+35i3
x3=-53-35i3
However, we have u2=x
So add 2 on both sides, and we get
u1=2(53-1)
u2=-53-2+35i3
u2=-53-2-35i3

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