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Question Number 38515 by Rio Mike last updated on 26/Jun/18

Question ; x^3 + x^3 = A) x^9 B) x^6 C) x^3 D) 1 Give a reason for your answer.

$$\:{Question}\:; \\ $$$${x}^{\mathrm{3}} \:+\:{x}^{\mathrm{3}} \:=\: \\ $$$$\left.{A}\right)\:{x}^{\mathrm{9}} \\ $$$$\left.{B}\right)\:{x}^{\mathrm{6}} \\ $$$$\left.{C}\right)\:{x}^{\mathrm{3}} \\ $$$$\left.{D}\right)\:\mathrm{1} \\ $$$${Give}\:{a}\:{reason}\:{for}\:{your}\:{answer}. \\ $$

Answered by MJS last updated on 26/Jun/18

x^3 +x^3 =x^9 ⇒ 2x^3 =x^9 ⇒ x=0 ∨ x^6 =2 ⇒ ⇒ x=0 ∨ x=±(2)^(1/6) ∨ x=−((2)^(1/6) /2)±i((2)^(1/6) /2)(√3) ∨ x=((2)^(1/6) /2)±i((2)^(1/6) /2)(√3) x^3 +x^3 =x^6 ⇒ 2x^3 =x^6 ⇒ x=0 ∨ x^3 =2 ⇒ ⇒ x=0 ∨ x=(2)^(1/3) ∨ x=−((2)^(1/3) /2)±i((2)^(1/3) /2)(√3) x^3 +x^3 =x^3 ⇒ x^3 =0 ⇒ x=0 x^3 +x^3 =1 ⇒ x^3 =(1/2) ⇒ x=((4)^(1/3) /2) ∨ x=−((4)^(1/3) /4)±i((4)^(1/3) /4)(√3)

$${x}^{\mathrm{3}} +{x}^{\mathrm{3}} ={x}^{\mathrm{9}} \:\Rightarrow\:\mathrm{2}{x}^{\mathrm{3}} ={x}^{\mathrm{9}} \:\Rightarrow\:{x}=\mathrm{0}\:\vee\:{x}^{\mathrm{6}} =\mathrm{2}\:\Rightarrow \\ $$$$\Rightarrow\:{x}=\mathrm{0}\:\vee\:{x}=\pm\sqrt[{\mathrm{6}}]{\mathrm{2}}\:\vee\:{x}=−\frac{\sqrt[{\mathrm{6}}]{\mathrm{2}}}{\mathrm{2}}\pm\mathrm{i}\frac{\sqrt[{\mathrm{6}}]{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\vee\:{x}=\frac{\sqrt[{\mathrm{6}}]{\mathrm{2}}}{\mathrm{2}}\pm\mathrm{i}\frac{\sqrt[{\mathrm{6}}]{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{3}} ={x}^{\mathrm{6}} \:\Rightarrow\:\mathrm{2}{x}^{\mathrm{3}} ={x}^{\mathrm{6}} \:\Rightarrow\:{x}=\mathrm{0}\:\vee\:{x}^{\mathrm{3}} =\mathrm{2}\:\Rightarrow \\ $$$$\Rightarrow\:{x}=\mathrm{0}\:\vee\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\vee\:{x}=−\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\mathrm{2}}\pm\mathrm{i}\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{3}} ={x}^{\mathrm{3}} \:\Rightarrow\:{x}^{\mathrm{3}} =\mathrm{0}\:\Rightarrow\:{x}=\mathrm{0} \\ $$$$ \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{3}} =\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{2}}\:\vee\:{x}=−\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{4}}\pm\mathrm{i}\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{4}}\sqrt{\mathrm{3}} \\ $$

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