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If-2-1-3-4-1-3-8-1-3-x-then-x-3-6x-2-6x-6-




Question Number 83941 by john santu last updated on 08/Mar/20
If ((2 ))^(1/3)  + (4)^(1/(3 ))  + ((8 ))^(1/(3 ))  = x   then x^3 −6x^2 +6x+6 = ?
$$\mathrm{If}\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{4}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{8}\:}\:=\:\mathrm{x}\: \\ $$$$\mathrm{then}\:\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{6}\:=\:? \\ $$
Answered by $@ty@m123 last updated on 08/Mar/20
^3 (√2)+^3 (√4)+2=x  ^3 (√2)+^3 (√4) =x−2  (^3 (√2)+^3 (√4))^3 =(x−2)^3   2+4+3.^3 (√8)(^3 (√2)+^3 (√4) )=(x−2)^3   6+3×2(^3 (√2)+^3 (√4) )=(x−2)^3   6+6(x−2)=(x−2)^3   6+6x−12=x^3 −8−6x^2 +12x  x^3 −6x^2 +6x+6=8
$$\:^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}+\mathrm{2}={x} \\ $$$$\:^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}\:={x}−\mathrm{2} \\ $$$$\left(\:^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}\right)^{\mathrm{3}} =\left({x}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\mathrm{2}+\mathrm{4}+\mathrm{3}.\:^{\mathrm{3}} \sqrt{\mathrm{8}}\left(\:^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}\:\right)=\left({x}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\mathrm{6}+\mathrm{3}×\mathrm{2}\left(\:^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}\:\right)=\left({x}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\mathrm{6}+\mathrm{6}\left({x}−\mathrm{2}\right)=\left({x}−\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\mathrm{6}+\mathrm{6}{x}−\mathrm{12}={x}^{\mathrm{3}} −\mathrm{8}−\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x} \\ $$$${x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{6}=\mathrm{8} \\ $$
Commented by john santu last updated on 08/Mar/20
good...
$$\mathrm{good}… \\ $$

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