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# help

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181
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Let a and b be nonzero complex numbers such that a^2 + ab + b^2 = 0. Evaluate $$\frac{a^9 + b^9}{(a + b)^9}.$$

Jan 29, 2019

#1
+5651
+2

I don't have a simple answer yet.

Brute force works.

Jan 29, 2019
#2
+28125
+2

As follows

Jan 30, 2019
#3
+22884
+8

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

$$\dfrac{a^9 + b^9}{(a + b)^9}.$$

$$\begin{array}{|lrcl|} \hline 1. & a^2+ab+b^2 &=& 0 \\ & \mathbf{a^2+b^2} & \mathbf{=}& \mathbf{ -ab } \\\\ 2. & (a^2+b^2)^2 &=& a^2b^2 \\ & a^4+2a^2b^2+b^4 &=& a^2b^2 \\ &\mathbf {a^4+b^4 } &\mathbf{=}&\mathbf {-a^2b^2} \\\\ 3. & (a^2+b^2)(a^4+b^4) &=& (-ab)(-a^2b^2) \\ & a^6+a^2b^4+b^2a^4+b^6&=&a^3b^3 \\ &a^6+b^6+a^2b^2(a^2+b^2) &=& a^3b^3 \\ &a^6+b^6+a^2b^2(-ab) &=& a^3b^3 \\ & \mathbf{ a^6+b^6-a^3b^3 }& \mathbf{=}& \mathbf{a^3b^3} \\\\ 4. & (a+b)^2 &=& a^2+2ab+b^2 \\ & (a+b)^2-ab &=& a^2+ab+b^2 \quad | \quad a^2+ab+b^2 =0 \\ & (a+b)^2-ab &=& 0\\ & \mathbf{ (a+b)^2 } &\mathbf{ =}& \mathbf{ ab} \\\\ 5. &\left( (a+b)^2\right)^4 &=& a^4b^4 \\ & \mathbf{ (a+b)^8} & \mathbf{ =}& \mathbf{ a^4b^4} \\ \hline \end{array}$$

$$\begin{array}{|rcl|} \hline && \mathbf{\dfrac{a^9+b^9}{(a+b)^9} } \\\\ &=& \dfrac{(a+b)(a^2-ab+b^2)(a^6-a^3b^3+b^6) } {(a+b)^8(a+b) } \\\\ &=& \dfrac{(a^2-ab+b^2)(a^6-a^3b^3+b^6) } {(a+b)^8 } \\\\ &=& \dfrac{(a^2+b^2-ab )(a^6-a^3b^3+b^6) } {(a+b)^8 } \\\\ &=& \dfrac{(-ab-ab )(a^6-a^3b^3+b^6) } {(a+b)^8 } \\\\ &=& \dfrac{(-2ab)(a^6-a^3b^3+b^6) } {(a+b)^8 } \\\\ &=& \dfrac{(-2ab)(a^3b^3) } {a^4b^4 } \\\\ &=& \dfrac{-2a^4b^4 } {a^4b^4 } \\\\ & \mathbf{=} & \mathbf{-2 }\\ \hline \end{array}$$

Jan 31, 2019