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# thanks complex numbers

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Let a and b be nonzero complex numbers such that a^2 + ab + b^2 = 0. Evaluate (a^9 + b^9)/(a + b)^9.

Mar 27, 2019

#1
+6192
+1

$$a^2 + ab + b^2 = 0\\ (a+b)^2 = a^2+2ab+b^2= ab\\ a = \dfrac{-b\pm \sqrt{-3b^2}}{2} = b e^{\pm i 2\pi/3}\\ a^9+b^9 = b^9(e^{\pm i 6\pi}+1) = 2b^9\\ (a+b)^9 =( (a+b)^2)^{9/2} = \\ (ab)^{9/2} =\left( b^2 e^{\pm i 2\pi/3}\right)^{9/2} = b^9e^{\pm i 3\pi} = -b^9\\ \dfrac{a^9+b^9}{(a+b)^9} = \dfrac{2b^9}{-b^9} = -2$$

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Mar 27, 2019
#2
+4580
+1

That's what I get too, is it right, ant101?

tertre  Mar 28, 2019
edited by tertre  Mar 28, 2019