x-1-x-x-4-3-x-2-3-x-1-3-x-2-1-x-2-prove-LHS-RHS-

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

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}}) = x^{\frac{1}{3}} (x^{2} - \frac{1}{x^{2}})

prove LHS = RHS

Commented by Socracious last updated on 24/Nov/22

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}}) = x^{\frac{1}{3}} (x^{2} - \frac{1}{x^{2}})

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}}) = x^{\frac{1}{3}} ((x + \frac{1}{x}) (x - \frac{1}{x}))

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}}) = (x - \frac{1}{x}) (x^{\frac{1}{3}} ∙ x^{1} + x^{\frac{1}{3}} ∙ x^{- 1})

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}}) = (x - \frac{1}{x}) (x^{\frac{1}{3} + 1} + x^{\frac{1}{3} - 1})

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}}) = (x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}})

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

L a s t t w o l i n e s : \frac{1}{3} - 1 = - \frac{2}{3}

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

y o u {c a n}^{'} t p r o v e, b e c a u s e L H S \neq R H S .

p l e a s e c h e c k y o u r t y p o .

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

w i t h x = 8

L H S = (8 - \frac{1}{8}) (16 + 4) = \frac{315}{2}

R H S = 2 (64 - \frac{1}{64}) = \frac{4095}{32}

\Rightarrow L H S \neq R H S

Commented by Mastermind last updated on 23/Nov/22

Thank

i^{'} ve been trying this question since \dots

it might be a mistake

thanks boss

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

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}}) = x^{\frac{1}{3}} (x^{2} - \frac{1}{x^{2}})

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{\frac{2}{3}}) = x^{\frac{1}{3}} (x - \frac{1}{x}) (x + \frac{1}{x})

x^{\frac{4}{3}} + x^{\frac{2}{3}} = x^{\frac{1}{3}} (x + \frac{1}{x})

x^{\frac{4}{3}} + x^{\frac{2}{3}} = x^{\frac{1}{3} + 1} + x^{\frac{1}{3} - 1}

x^{\frac{4}{3}} + x^{\frac{2}{3}} = x^{\frac{4}{3}} + x^{- \frac{2}{3}}

x^{\frac{2}{3}} = x^{- \frac{2}{3}}

Correct only when x = 1

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

Seems typo in your question . The

following is correct version .

(x - \frac{1}{x}) (x^{\frac{4}{3}} + x^{- \frac{2}{3}}) = x^{\frac{1}{3}} (x^{2} - \frac{1}{x^{2}})

Answered by SEKRET last updated on 23/Nov/22

(x - \frac{1}{x}) \cdot x^{\frac{1}{3}} \cdot (x + x^{\frac{1}{3}}) = x^{\frac{1}{3}} \cdot (x - \frac{1}{x}) \cdot (x + \frac{1}{x})

x \neq 0 x - \frac{1}{x} = 0 x = 1 x = - 1

x + x^{\frac{1}{3}} = x + \frac{1}{x} x^{\frac{4}{3}} = 1 x = 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 ?

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