Question 29244: sixth root of (-1+sqrt3i)^3
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! sixth root of (-1+sqrt3i)^3
We have (a+b)^3 = (a)^3+(b)^3+3ab(a+b)
Here a= (-1) therefore a^3 = (-1)^3 = -1
b = (i sqrt3) therefore b^3 = (isqrt3)^3 = (i)^3(sqrt3)^3= (i^2Xi)X(3)X(sqrt3)= -3i(sqrt3)
3ab(a+b)= 3a^2b + 3ab^2
= 3X(-1)^2X(i sqrt3) + 3X(-1)X(i sqrt3)^2
=3X1X(i sqrt3)-3X(i^2)X(3)
=3i(sqrt3)-3X(-1)X3
=3i(sqrt3)+9
Therefore
(-1+sqrt3i)^3= (-1)^3 +( isqrt3)^3+3X(-1)^2X(i sqrt3) + 3X(-1)X(i sqrt3)^2
= -1-3i(sqrt3)+3i(sqrt3)+9
=-1+0+9 = 8
Sixth root of (-1+sqrt3i)^3 = sixth root of (8)
= (8)^(1/6)
= [(2^3)]^(1/6)
=(2)^(3X1/6)
=(2)^(1/2)
=sqrt(2)
Answer: sqrt(2)
Note: If you are in a class which has not yet introduced you to the concept of finding the nth roots of a complex number using trigonometric representation for a complex number and DeMoivre's Theorem, the above answer will do. Please ask for the theory of the above mentioned concept and examples if you are in a high school class.
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