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Question Number 181245 by Mastermind last updated on 23/Nov/22

(x−(1/x))(x^(4/3) +x^(2/3) )=x^(1/3) (x^2 −(1/x^2 )) prove LHS=RHS

$$\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$$\mathrm{prove}\:\mathrm{LHS}=\mathrm{RHS} \\ $$

Commented by Socracious last updated on 24/Nov/22

(x−(1/x))(x^(4/3) +x^(2/3) )=x^(1/3) (x^2 −(1/x^2 )) (x−(1/x))(x^(4/3) +x^(2/3) )=x^(1/3) ((x+(1/x))(x−(1/x))) (x−(1/x))(x^(4/3) +x^(2/3) )=(x−(1/x))(x^(1/3) •x^1 +x^(1/3) •x^(−1) ) (x−(1/x))(x^(4/3) +x^(2/3) )=(x−(1/x))(x^((1/3)+1) +x^((1/3)−1) ) (x−(1/x))(x^(4/3) +x^(2/3) )=(x−(1/x))(x^(4/3) +x^(2/3) )

$$\:\:\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}^{\frac{\mathrm{4}}{\mathrm{3}}} +\boldsymbol{\mathrm{x}}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}^{\frac{\mathrm{4}}{\mathrm{3}}} +\boldsymbol{\mathrm{x}}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\right) \\ $$$$\:\:\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}^{\frac{\mathrm{4}}{\mathrm{3}}} +\boldsymbol{\mathrm{x}}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \bullet\boldsymbol{\mathrm{x}}^{\mathrm{1}} +\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \bullet\boldsymbol{\mathrm{x}}^{−\mathrm{1}} \right) \\ $$$$\:\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}^{\frac{\mathrm{4}}{\mathrm{3}}} +\boldsymbol{\mathrm{x}}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}} +\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \right) \\ $$$$\:\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}^{\frac{\mathrm{4}}{\mathrm{3}}} +\boldsymbol{\mathrm{x}}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\left(\boldsymbol{\mathrm{x}}^{\frac{\mathrm{4}}{\mathrm{3}}} +\boldsymbol{\mathrm{x}}^{\frac{\mathrm{2}}{\mathrm{3}}} \right) \\ $$

Commented by Rasheed.Sindhi last updated on 25/Nov/22

Last two lines: (1/3)−1=−(2/3)

$${Last}\:{two}\:{lines}:\:\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Commented by mr W last updated on 23/Nov/22

you can′t prove, because LHS≠RHS. please check your typo.

$${you}\:{can}'{t}\:{prove},\:{because}\:{LHS}\neq{RHS}. \\ $$$${please}\:{check}\:{your}\:{typo}. \\ $$

Commented by mr W last updated on 23/Nov/22

with x=8 LHS=(8−(1/8))(16+4)=((315)/2) RHS=2(64−(1/(64)))=((4095)/(32)) ⇒LHS≠RHS

$${with}\:{x}=\mathrm{8} \\ $$$${LHS}=\left(\mathrm{8}−\frac{\mathrm{1}}{\mathrm{8}}\right)\left(\mathrm{16}+\mathrm{4}\right)=\frac{\mathrm{315}}{\mathrm{2}} \\ $$$${RHS}=\mathrm{2}\left(\mathrm{64}−\frac{\mathrm{1}}{\mathrm{64}}\right)=\frac{\mathrm{4095}}{\mathrm{32}} \\ $$$$\Rightarrow{LHS}\neq{RHS} \\ $$

Commented by Mastermind last updated on 23/Nov/22

Thank i′ve been trying this question since ... it might be a mistake thanks boss

$$\mathrm{Thank} \\ $$$$\mathrm{i}'\mathrm{ve}\:\mathrm{been}\:\mathrm{trying}\:\mathrm{this}\:\mathrm{question}\:\mathrm{since}\:... \\ $$$$\mathrm{it}\:\mathrm{might}\:\mathrm{be}\:\mathrm{a}\:\mathrm{mistake} \\ $$$$\mathrm{thanks}\:\mathrm{boss} \\ $$$$ \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 23/Nov/22

(x−(1/x))(x^(4/3) +x^(2/3) )=x^(1/3) (x^2 −(1/x^2 )) (x−(1/x))(x^(4/3) +x^(2/3) )=x^(1/3) (x−(1/x))(x+(1/x)) x^(4/3) +x^(2/3) =x^(1/3) (x+(1/x)) x^(4/3) +x^(2/3) =x^((1/3)+1) +x^((1/3)−1) x^(4/3) +x^(2/3) =x^(4/3) +x^(−(2/3)) x^(2/3) =x^(−(2/3)) Correct only when x=1

$$\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$\cancel{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)}\left(\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \cancel{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}} +\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \\ $$$$\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\mathrm{Correct}\:\mathrm{only}\:\mathrm{when}\:\mathrm{x}=\mathrm{1} \\ $$

Commented by Rasheed.Sindhi last updated on 23/Nov/22

Seems typo in your question.The following is correct version. (x−(1/x))(x^(4/3) +x^(−(2/3)) )=x^(1/3) (x^2 −(1/x^2 ))

$$\mathrm{Seems}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{your}\:\mathrm{question}.\mathrm{The} \\ $$$$\mathrm{following}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{version}. \\ $$$$\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} \right)=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$

Answered by SEKRET last updated on 23/Nov/22

(x−(1/x))∙x^(1/3) ∙(x+x^(1/3) )=x^(1/3) ∙(x−(1/x))∙(x+(1/x)) x≠0 x−(1/x)=0 x = 1 x= −1 x+x^(1/3) =x+(1/x) x^(4/3) =1 x=1

$$\:\:\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\centerdot\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \centerdot\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)=\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \centerdot\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)\centerdot\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{x}}\neq\mathrm{0}\:\:\:\:\:\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}=\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{x}}\:=\:\mathrm{1}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\:−\mathrm{1} \\ $$$$\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} =\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\frac{\mathrm{4}}{\mathrm{3}}} =\mathrm{1}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{1} \\ $$$$ \\ $$

Commented by Frix last updated on 23/Nov/22

It makes no sense to solve when a proof is requested. Or would you welcome a piece of bread when you are thirsty?

$$\mathrm{It}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{when}\:\mathrm{a}\:\mathrm{proof}\:\mathrm{is} \\ $$$$\mathrm{requested}.\:\mathrm{Or}\:\mathrm{would}\:\mathrm{you}\:\mathrm{welcome}\:\mathrm{a}\:\mathrm{piece} \\ $$$$\mathrm{of}\:\mathrm{bread}\:\mathrm{when}\:\mathrm{you}\:\mathrm{are}\:\mathrm{thirsty}? \\ $$

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