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Question Number 7610 by Tawakalitu. last updated on 06/Sep/16

obtain the value of (a^(1/2) +b^(1/2) +c^(1/2) )(a^(1/2) −b^(1/2) +c^(1/2) )(a^(1/2) −b^(1/2) +c^(1/2) )(b^(1/2) +c^(1/2) −a^(1/2) )

$${obtain}\:{the}\:{value}\:{of}\: \\ $$$$\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} −{a}^{\mathrm{1}/\mathrm{2}} \right)\: \\ $$

Commented by Rasheed Soomro last updated on 06/Sep/16

Perhaps you mean (a^(1/2) +b^(1/2) +c^(1/2) )(a^(1/2) −b^(1/2) +c^(1/2) )(a^(1/2) +b^(1/2) −c^(1/2) )(b^(1/2) +c^(1/2) −a^(1/2) )

$${Perhaps}\:{you}\:{mean} \\ $$$$\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} −{c}^{\mathrm{1}/\mathrm{2}} \right)\left({b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} −{a}^{\mathrm{1}/\mathrm{2}} \right)\: \\ $$

Commented by Tawakalitu. last updated on 06/Sep/16

Alright sir.. thank you. is like that.

$${Alright}\:{sir}..\:{thank}\:{you}.\:{is}\:{like}\:{that}. \\ $$

Answered by Rasheed Soomro last updated on 06/Sep/16

(a^(1/2) +b^(1/2) +c^(1/2) )(a^(1/2) −b^(1/2) +c^(1/2) )(a^(1/2) +b^(1/2) −c^(1/2) )(b^(1/2) +c^(1/2) −a^(1/2) ) Let a^(1/2) =p , b^(1/2) =q , c^(1/2) =r (p+q+r)(p−q+r)(p+q−r)(−p+q+r) {(p+r)+q}{(p+r)−q}{q+(p−r)}{q−(p−r)} {(p+r)^2 −q^2 }{q^2 −(p−r)^2 } (p^2 +2pr+r^2 −q^2 )(q^2 −p^2 +2pr−r^2 ) {2pr+(p^2 −q^2 +r^2 )}{2pr−(p^2 −q^2 +r^2 )} (2pr)^2 −(p^2 −q^2 +r^2 )^2 4p^2 r^2 −(p^4 +q^4 +r^4 −2p^2 q^2 −2q^2 r^2 +2p^2 r^2 ) 2p^2 r^2 −p^4 −q^4 −r^4 +2p^2 q^2 +2q^2 r^2 p=a^(1/2) ⇒p^2 =a⇒p^4 =a^2 q=b^(1/2) ⇒q^2 =b⇒q^4 =b^2 r=c^(1/2) ⇒r^2 =c⇒r^4 =c^2 2p^2 r^2 −p^4 −q^4 −r^4 +2p^2 q^2 +2q^2 r^2 =2ac−a^2 −b^2 −c^2 +2ab+2bc =−(a^2 +b^2 +c^2 −2ab−2bc−2ca)

$$\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} −{c}^{\mathrm{1}/\mathrm{2}} \right)\left({b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} −{a}^{\mathrm{1}/\mathrm{2}} \right)\: \\ $$$${Let}\:\:{a}^{\mathrm{1}/\mathrm{2}} ={p}\:,\:{b}^{\mathrm{1}/\mathrm{2}} ={q}\:,\:{c}^{\mathrm{1}/\mathrm{2}} ={r} \\ $$$$\left({p}+{q}+{r}\right)\left({p}−{q}+{r}\right)\left({p}+{q}−{r}\right)\left(−{p}+{q}+{r}\right) \\ $$$$\left\{\left({p}+{r}\right)+{q}\right\}\left\{\left({p}+{r}\right)−{q}\right\}\left\{{q}+\left({p}−{r}\right)\right\}\left\{{q}−\left({p}−{r}\right)\right\} \\ $$$$\left\{\left({p}+{r}\right)^{\mathrm{2}} −{q}^{\mathrm{2}} \right\}\left\{{q}^{\mathrm{2}} −\left({p}−{r}\right)^{\mathrm{2}} \right\} \\ $$$$\left({p}^{\mathrm{2}} +\mathrm{2}{pr}+{r}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)\left({q}^{\mathrm{2}} −{p}^{\mathrm{2}} +\mathrm{2}{pr}−{r}^{\mathrm{2}} \right) \\ $$$$\left\{\mathrm{2}{pr}+\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)\right\}\left\{\mathrm{2}{pr}−\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)\right\} \\ $$$$\left(\mathrm{2}{pr}\right)^{\mathrm{2}} −\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{4}{p}^{\mathrm{2}} {r}^{\mathrm{2}} −\left({p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{r}^{\mathrm{4}} −\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{2}{p}^{\mathrm{2}} {r}^{\mathrm{2}} \right) \\ $$$$\mathrm{2}{p}^{\mathrm{2}} {r}^{\mathrm{2}} −{p}^{\mathrm{4}} −{q}^{\mathrm{4}} −{r}^{\mathrm{4}} +\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$${p}={a}^{\mathrm{1}/\mathrm{2}} \Rightarrow{p}^{\mathrm{2}} ={a}\Rightarrow{p}^{\mathrm{4}} ={a}^{\mathrm{2}} \\ $$$${q}={b}^{\mathrm{1}/\mathrm{2}} \Rightarrow{q}^{\mathrm{2}} ={b}\Rightarrow{q}^{\mathrm{4}} ={b}^{\mathrm{2}} \\ $$$${r}={c}^{\mathrm{1}/\mathrm{2}} \Rightarrow{r}^{\mathrm{2}} ={c}\Rightarrow{r}^{\mathrm{4}} ={c}^{\mathrm{2}} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} {r}^{\mathrm{2}} −{p}^{\mathrm{4}} −{q}^{\mathrm{4}} −{r}^{\mathrm{4}} +\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{ac}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2bc} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{ab}−\mathrm{2}{bc}−\mathrm{2}{ca}\right) \\ $$

Commented by Tawakalitu. last updated on 06/Sep/16

wow, thank you sir. God bless you

$${wow},\:{thank}\:{you}\:{sir}.\:{God}\:{bless}\:{you} \\ $$

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