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Question-92831

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Question Number 92831 by unknown last updated on 09/May/20
Commented by unknown last updated on 09/May/20
Solve the equation (p,q,r) if the solution are real number
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\left({p},{q},{r}\right)\:\mathrm{if}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{are}\:\mathrm{real}\:\mathrm{number} \\ $$
Commented by unknown last updated on 09/May/20
Solve the equation (p,q,r) if the solution are real number
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\left({p},{q},{r}\right)\:\mathrm{if}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{are}\:\mathrm{real}\:\mathrm{number} \\ $$$$ \\ $$
Commented by i jagooll last updated on 09/May/20
((p)^(1/(3 )) +(q)^(1/(3 )) +(r)^(1/(3 )) )^3 = (√2)−1 p+q+r+3((p)^(1/(3 )) +(q)^(1/(3 )) )((p)^(1/(3 )) +(r)^(1/(3 )) )((q)^(1/(3 )) +(r)^(1/(3 )) ) = (√2)−1 ⇒p+q+r = −1 ⇒3((p)^(1/(3 )) +(q)^(1/(3 )) )((p)^(1/(3 )) +(r)^(1/(3 )) )((q)^(1/(3 )) +(r)^(1/(3 )) )=(√2)
$$\left(\sqrt[{\mathrm{3}\:\:}]{\mathrm{p}}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{q}}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{r}}\:\right)^{\mathrm{3}} \:=\:\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{3}\left(\sqrt[{\mathrm{3}\:\:}]{\mathrm{p}}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{q}}\right)\left(\sqrt[{\mathrm{3}\:\:}]{\mathrm{p}}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{r}}\right)\left(\sqrt[{\mathrm{3}\:}]{\mathrm{q}}\:+\sqrt[{\mathrm{3}\:\:}]{\mathrm{r}}\right)\:=\:\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{p}+\mathrm{q}+\mathrm{r}\:=\:−\mathrm{1}\: \\ $$$$\Rightarrow\mathrm{3}\left(\sqrt[{\mathrm{3}\:}]{\mathrm{p}}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{q}}\right)\left(\sqrt[{\mathrm{3}\:}]{\mathrm{p}}\:+\sqrt[{\mathrm{3}\:\:}]{\mathrm{r}}\right)\left(\sqrt[{\mathrm{3}\:\:}]{\mathrm{q}}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{r}}\:\right)=\sqrt{\mathrm{2}} \\ $$
Commented by unknown last updated on 09/May/20
so the solution (p,q,r)(?
$${so}\:{the}\:{solution}\:\left({p},{q},{r}\right)\left(?\right. \\ $$$$ \\ $$
Commented by prakash jain last updated on 12/May/20
Jagooll from step 2 you cannot infer p+q+r=−1 when p,q,r ∈R
$$\mathrm{Jagooll}\:\mathrm{from}\:\mathrm{step}\:\mathrm{2}\: \\ $$$$\mathrm{you}\:\mathrm{cannot}\:\mathrm{infer} \\ $$$${p}+{q}+{r}=−\mathrm{1}\:\mathrm{when}\:{p},{q},{r}\:\in\mathbb{R} \\ $$
Answered by prakash jain last updated on 12/May/20
(p)^(1/3) +(q)^(1/3) +(r)^(1/3) =(((√2)−1))^(1/3) Given any value for p and q there exists a solution for r for example p=1,q=1 r=((((√2)−1))^(1/3) −2)^3 There are infinte number of solutions. Please recheck question
$$\sqrt[{\mathrm{3}}]{{p}}+\sqrt[{\mathrm{3}}]{{q}}+\sqrt[{\mathrm{3}}]{{r}}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$\mathrm{Given}\:\mathrm{any}\:\mathrm{value}\:\mathrm{for}\:{p}\:\mathrm{and}\:{q} \\ $$$$\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{for}\:{r} \\ $$$$\mathrm{for}\:\mathrm{example} \\ $$$${p}=\mathrm{1},{q}=\mathrm{1} \\ $$$${r}=\left(\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}−\mathrm{1}}\:−\mathrm{2}\right)^{\mathrm{3}} \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{infinte}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}. \\ $$$$\mathrm{Please}\:\mathrm{recheck}\:\mathrm{question} \\ $$

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