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Question Number 202447 by maths_plus last updated on 26/Dec/23
rationnalise le denominateur de   x = (((2)^(1/(3 )) −1)/(1−^3 (√2)+^3 (√4)))
$$\mathrm{rationnalise}\:\mathrm{le}\:\mathrm{denominateur}\:\mathrm{de}\: \\ $$$$\mathrm{x}\:=\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}−\mathrm{1}}{\mathrm{1}−^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}} \\ $$
Answered by cortano12 last updated on 27/Dec/23
 a^3 +b^3  = (a+b)(a^2 −ab+b^2 )   1^3 +((2)^(1/3) )^3 = (1+(2)^(1/3) )(1−(2)^(1/3) +(4)^(1/3)  )   (1/3) = (1/((1+(2)^(1/3)  )(1−(2)^(1/3)  +(4)^(1/3)  )))   ⇒x = (2)^(1/3) −1 .(((2)^(1/3) +1)/3)
$$\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} \:=\:\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right) \\ $$$$\:\mathrm{1}^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{3}} =\:\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}\:\right) \\ $$$$\:\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\right)\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{4}}\:\right)} \\ $$$$\:\Rightarrow\mathrm{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\:.\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}}{\mathrm{3}} \\ $$
Commented by Frix last updated on 27/Dec/23
Yes but x≠(2)^(1/3) −1.(((2)^(1/3) +1)/3)  x=((2)^(1/3) −1)(((2)^(1/3) +1)/3)=−((1−(4)^(1/3) )/3)
$$\mathrm{Yes}\:\mathrm{but}\:{x}\neq\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}.\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}}{\mathrm{3}} \\ $$$${x}=\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}}{\mathrm{3}}=−\frac{\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{3}} \\ $$
Commented by cortano12 last updated on 27/Dec/23
 why?
$$\:\mathrm{why}? \\ $$
Commented by Frix last updated on 27/Dec/23
Sorry there was a typo.  But essentially you need brackets.  a+b×(c/d)≠(a+b)×(c/d)
$$\mathrm{Sorry}\:\mathrm{there}\:\mathrm{was}\:\mathrm{a}\:\mathrm{typo}. \\ $$$$\mathrm{But}\:\mathrm{essentially}\:\mathrm{you}\:\mathrm{need}\:\mathrm{brackets}. \\ $$$${a}+{b}×\frac{{c}}{{d}}\neq\left({a}+{b}\right)×\frac{{c}}{{d}} \\ $$
Answered by MATHEMATICSAM last updated on 27/Dec/23
1^3  + ((2)^(1/3) )^3  = ((2)^(1/3)  + 1)(1 − (2)^(1/3)  + (4)^(1/3) )  Now x = ((((2)^(1/3)  −1)((2)^(1/3)  + 1))/(((2)^(1/3)  + 1)(1 − (2)^(1/3)  + (4)^(1/3) )))  = ((((2)^(1/3) )^2  − 1)/(1 + ((2)^(1/3) )^3 )) = (((4)^(1/3)  − 1)/(1 + 2)) = (((4)^(1/3)  − 1)/3) (Ans)
$$\mathrm{1}^{\mathrm{3}} \:+\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{3}} \:=\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{1}\right)\left(\mathrm{1}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right) \\ $$$$\mathrm{Now}\:{x}\:=\:\frac{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{1}\right)}{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{1}\right)\left(\mathrm{1}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)} \\ $$$$=\:\frac{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{2}} \:−\:\mathrm{1}}{\mathrm{1}\:+\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{3}} }\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}\:−\:\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}\:−\:\mathrm{1}}{\mathrm{3}}\:\left(\mathrm{Ans}\right) \\ $$

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