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Question Number 21877 by FilupS last updated on 06/Oct/17
x=^3 (√(7+5(√2)))+^3 (√(7−5(√2)))     1. According to a video, x=2     2. According to WolframAlpha,             x≈0.2071+0.3587i  for “principal root”  and   x=2(√2)  for “real-valued root”     3. According to google, x≈2.8284     Please help and explain!   :)
$${x}=^{\mathrm{3}} \sqrt{\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}}+^{\mathrm{3}} \sqrt{\mathrm{7}−\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$$\: \\ $$$$\mathrm{1}.\:\mathrm{According}\:\mathrm{to}\:\mathrm{a}\:\mathrm{video},\:{x}=\mathrm{2} \\ $$$$\: \\ $$$$\mathrm{2}.\:\mathrm{According}\:\mathrm{to}\:\mathrm{WolframAlpha}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}\approx\mathrm{0}.\mathrm{2071}+\mathrm{0}.\mathrm{3587}{i}\:\:\mathrm{for}\:“\mathrm{principal}\:\mathrm{root}'' \\ $$$$\mathrm{and}\:\:\:{x}=\mathrm{2}\sqrt{\mathrm{2}}\:\:\mathrm{for}\:“\mathrm{real}-\mathrm{valued}\:\mathrm{root}'' \\ $$$$\: \\ $$$$\mathrm{3}.\:\mathrm{According}\:\mathrm{to}\:\mathrm{google},\:{x}\approx\mathrm{2}.\mathrm{8284} \\ $$$$\: \\ $$$$\left.\mathrm{Please}\:\mathrm{help}\:\mathrm{and}\:\mathrm{explain}!\:\:\::\right) \\ $$
Commented by Joel577 last updated on 06/Oct/17
where′s the video?
$${where}'{s}\:{the}\:{video}? \\ $$
Commented by Joel577 last updated on 06/Oct/17
Commented by FilupS last updated on 06/Oct/17
Answered by mrW1 last updated on 06/Oct/17
a=^3 (√(7+5(√2)))  b=^3 (√(7−5(√2)))  x=a+b=real  a^3 +b^3 =14  ab=^3 (√((7+5(√2))(7−5(√2))))=^3 (√(49−50))=−1  a^3 +b^3 =(a+b)(a^2 −ab+b^2 )  a^3 +b^3 =(a+b)(a^2 +2ab+b^2 −3ab)  a^3 +b^3 =(a+b)[(a+b)^2 −3ab]  ⇒14=x(x^2 +3)  ⇒x^3 +3x−14=0  x^3 −2x^2 +2x^2 −4x+7x−14=0  (x−2)(x^2 +2x+7)=0  ⇒x−2=0  ⇒x=2
$$\mathrm{a}=^{\mathrm{3}} \sqrt{\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{b}=^{\mathrm{3}} \sqrt{\mathrm{7}−\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{x}=\mathrm{a}+\mathrm{b}=\mathrm{real} \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{14} \\ $$$$\mathrm{ab}=^{\mathrm{3}} \sqrt{\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right)\left(\mathrm{7}−\mathrm{5}\sqrt{\mathrm{2}}\right)}=^{\mathrm{3}} \sqrt{\mathrm{49}−\mathrm{50}}=−\mathrm{1} \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{2}} +\mathrm{2ab}+\mathrm{b}^{\mathrm{2}} −\mathrm{3ab}\right) \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\left(\mathrm{a}+\mathrm{b}\right)\left[\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{3ab}\right] \\ $$$$\Rightarrow\mathrm{14}=\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{3}} +\mathrm{3x}−\mathrm{14}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{7x}−\mathrm{14}=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2} \\ $$
Answered by myintkhaing last updated on 06/Oct/17

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