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Question-63642

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Question Number 63642 by bshahid010@gmail.com last updated on 06/Jul/19
Commented by kaivan.ahmadi last updated on 06/Jul/19
(x+2)(x^2 −2x+4)=0⇒ x=−2⇒γ=−2⇒γ^2 =4 α,β are the roots of x^2 −2x+4=0 the equation is (x−α^2 )(x−β^2 )(x−4)=0 (x^2 −(α^2 +β^2 )x+α^2 β^2 )(x−4)=0 but α+β=2,αβ=4 α^2 +β^2 =(α+β)^2 −2αβ=4−8=−4 α^2 β^2 =(αβ)^2 =16 so the equation is (x^2 +4x+16)(x−4)=0⇒ x^3 −4x^2 +4x^2 −16x+16x−64=0⇒ x^3 −64=0
$$\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}\right)=\mathrm{0}\Rightarrow \\ $$$${x}=−\mathrm{2}\Rightarrow\gamma=−\mathrm{2}\Rightarrow\gamma^{\mathrm{2}} =\mathrm{4} \\ $$$$\alpha,\beta\:{are}\:{the}\:{roots}\:{of}\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$${the}\:{equation}\:{is} \\ $$$$\left({x}−\alpha^{\mathrm{2}} \right)\left({x}−\beta^{\mathrm{2}} \right)\left({x}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right){x}+\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \right)\left({x}−\mathrm{4}\right)=\mathrm{0} \\ $$$${but}\: \\ $$$$\alpha+\beta=\mathrm{2},\alpha\beta=\mathrm{4} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta=\mathrm{4}−\mathrm{8}=−\mathrm{4} \\ $$$$\alpha^{\mathrm{2}} \beta^{\mathrm{2}} =\left(\alpha\beta\right)^{\mathrm{2}} =\mathrm{16} \\ $$$${so}\:{the}\:{equation}\:{is} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{16}\right)\left({x}−\mathrm{4}\right)=\mathrm{0}\Rightarrow \\ $$$${x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{16}{x}−\mathrm{64}=\mathrm{0}\Rightarrow \\ $$$${x}^{\mathrm{3}} −\mathrm{64}=\mathrm{0} \\ $$$$ \\ $$
Answered by MJS last updated on 06/Jul/19
x^3 −8^2 =0
$${x}^{\mathrm{3}} −\mathrm{8}^{\mathrm{2}} =\mathrm{0} \\ $$
Answered by MJS last updated on 06/Jul/19
x^3 =−8 x_1 =−2 ⇒ x_1 ^2 =4 x_2 =2e^(i(π/3)) ⇒ x_2 ^2 =4e^(i((2π)/3)) x_3 =2e^(−i(π/3)) ⇒ x_3 ^2 =4e^(−i((2π)/3)) we know that x_1 ^3 =64 ⇒ the equation is x^3 =64 to be sure: (x−4)(x−4e^(i((2π)/3)) )(x−4e^(−i((2π)/3)) )= =(x−4)(x^2 −4x(e^(i((2π)/3)) +e^(−i((2π)/3)) )+16)= =(x−4)(x^2 −4x(−1)+16)= =(x−4)(x^2 +4x+16)=x^3 −64
$${x}^{\mathrm{3}} =−\mathrm{8} \\ $$$${x}_{\mathrm{1}} =−\mathrm{2}\:\Rightarrow\:{x}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{4} \\ $$$${x}_{\mathrm{2}} =\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \:\Rightarrow\:{x}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{4e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${x}_{\mathrm{3}} =\mathrm{2e}^{−\mathrm{i}\frac{\pi}{\mathrm{3}}} \:\Rightarrow\:{x}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{4e}^{−\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:{x}_{\mathrm{1}} ^{\mathrm{3}} =\mathrm{64}\:\Rightarrow\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is} \\ $$$${x}^{\mathrm{3}} =\mathrm{64} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{sure}: \\ $$$$\left({x}−\mathrm{4}\right)\left({x}−\mathrm{4e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)\left({x}−\mathrm{4e}^{−\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)= \\ $$$$=\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} −\mathrm{4}{x}\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} +\mathrm{e}^{−\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)+\mathrm{16}\right)= \\ $$$$=\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} −\mathrm{4}{x}\left(−\mathrm{1}\right)+\mathrm{16}\right)= \\ $$$$=\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{16}\right)={x}^{\mathrm{3}} −\mathrm{64} \\ $$

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