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

$$(\mathrm{x}-\frac{1}{\mathrm{x}})({\mathrm{x}}^{\frac{4}{3}}+{\mathrm{x}}^{\frac{2}{3}})={\mathrm{x}}^{\frac{1}{3}}({\mathrm{x}}^2-\frac{1}{{\mathrm{x}}^2})$$$$$$$$\mathrm{prove}\mathrm{LHS}=\mathrm{RHS}$$

Commented by Socracious last updated on 24/Nov/22

$$(\mathbf{x}-\frac{1}{\mathbf{x}})({\mathbf{x}}^{\frac{4}{3}}+{\mathbf{x}}^{\frac{2}{3}})={\mathbf{x}}^{\frac{1}{3}}({\mathbf{x}}^2-\frac{1}{{\mathbf{x}}^2})$$$$(\mathbf{x}-\frac{1}{\mathbf{x}})({\mathbf{x}}^{\frac{4}{3}}+{\mathbf{x}}^{\frac{2}{3}})={\mathbf{x}}^{\frac{1}{3}}\left((\mathbf{x}+\frac{1}{\mathbf{x}})(\mathbf{x}-\frac{1}{\mathbf{x}})\right)$$$$(\mathbf{x}-\frac{1}{\mathbf{x}})({\mathbf{x}}^{\frac{4}{3}}+{\mathbf{x}}^{\frac{2}{3}})=(\mathbf{x}-\frac{1}{\mathbf{x}})({\mathbf{x}}^{\frac{1}{3}}\bullet {\mathbf{x}}^1+{\mathbf{x}}^{\frac{1}{3}}\bullet {\mathbf{x}}^{-1})$$$$(\mathbf{x}-\frac{1}{\mathbf{x}})({\mathbf{x}}^{\frac{4}{3}}+{\mathbf{x}}^{\frac{2}{3}})=(\mathbf{x}-\frac{1}{\mathbf{x}})({\mathbf{x}}^{\frac{1}{3}+1}+{\mathbf{x}}^{\frac{1}{3}-1})$$$$(\mathbf{x}-\frac{1}{\mathbf{x}})({\mathbf{x}}^{\frac{4}{3}}+{\mathbf{x}}^{\frac{2}{3}})=(\mathbf{x}-\frac{1}{\mathbf{x}})({\mathbf{x}}^{\frac{4}{3}}+{\mathbf{x}}^{\frac{2}{3}})$$

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

$$Lasttwolines:\frac{1}{3}-1=-\frac{2}{3}$$

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

$$you{can}^{\prime }tprove,becauseLHS\ne RHS.$$$$pleasecheckyourtypo.$$

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

$$withx=8$$$$LHS=(8-\frac{1}{8})(16+4)=\frac{315}{2}$$$$RHS=2(64-\frac{1}{64})=\frac{4095}{32}$$$$\Rightarrow LHS\ne RHS$$

Commented by Mastermind last updated on 23/Nov/22

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

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

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

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

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

Answered by SEKRET last updated on 23/Nov/22

$$(\mathbf{x}-\frac{1}{\mathbf{x}})\cdot {\mathbf{x}}^{\frac{1}{3}}\cdot (\mathbf{x}+{\mathbf{x}}^{\frac{1}{3}})={\mathbf{x}}^{\frac{1}{3}}\cdot (\mathbf{x}-\frac{1}{\mathbf{x}})\cdot (\mathbf{x}+\frac{1}{\mathbf{x}})$$$$\mathbf{x}\ne 0\mathbf{x}-\frac{1}{\mathbf{x}}=0\mathbf{x}=1\mathbf{x}=-1$$$$\mathbf{x}+{\mathbf{x}}^{\frac{1}{3}}=\mathbf{x}+\frac{1}{\mathbf{x}}{\mathbf{x}}^{\frac{4}{3}}=1\mathbf{x}=1$$$$$$

Commented by Frix last updated on 23/Nov/22

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

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