Question Number 7610 by Tawakalitu. last updated on 06/Sep/16
$${obtain}\:{the}\:{value}\:{of}\: \\ $$$$\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} −{a}^{\mathrm{1}/\mathrm{2}} \right)\: \\ $$
Commented by Rasheed Soomro last updated on 06/Sep/16
$${Perhaps}\:{you}\:{mean} \\ $$$$\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} −{c}^{\mathrm{1}/\mathrm{2}} \right)\left({b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} −{a}^{\mathrm{1}/\mathrm{2}} \right)\: \\ $$
Commented by Tawakalitu. last updated on 06/Sep/16
$${Alright}\:{sir}..\:{thank}\:{you}.\:{is}\:{like}\:{that}. \\ $$
Answered by Rasheed Soomro last updated on 06/Sep/16
$$\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} −{c}^{\mathrm{1}/\mathrm{2}} \right)\left({b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} −{a}^{\mathrm{1}/\mathrm{2}} \right)\: \\ $$$${Let}\:\:{a}^{\mathrm{1}/\mathrm{2}} ={p}\:,\:{b}^{\mathrm{1}/\mathrm{2}} ={q}\:,\:{c}^{\mathrm{1}/\mathrm{2}} ={r} \\ $$$$\left({p}+{q}+{r}\right)\left({p}−{q}+{r}\right)\left({p}+{q}−{r}\right)\left(−{p}+{q}+{r}\right) \\ $$$$\left\{\left({p}+{r}\right)+{q}\right\}\left\{\left({p}+{r}\right)−{q}\right\}\left\{{q}+\left({p}−{r}\right)\right\}\left\{{q}−\left({p}−{r}\right)\right\} \\ $$$$\left\{\left({p}+{r}\right)^{\mathrm{2}} −{q}^{\mathrm{2}} \right\}\left\{{q}^{\mathrm{2}} −\left({p}−{r}\right)^{\mathrm{2}} \right\} \\ $$$$\left({p}^{\mathrm{2}} +\mathrm{2}{pr}+{r}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)\left({q}^{\mathrm{2}} −{p}^{\mathrm{2}} +\mathrm{2}{pr}−{r}^{\mathrm{2}} \right) \\ $$$$\left\{\mathrm{2}{pr}+\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)\right\}\left\{\mathrm{2}{pr}−\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)\right\} \\ $$$$\left(\mathrm{2}{pr}\right)^{\mathrm{2}} −\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{4}{p}^{\mathrm{2}} {r}^{\mathrm{2}} −\left({p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{r}^{\mathrm{4}} −\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{2}{p}^{\mathrm{2}} {r}^{\mathrm{2}} \right) \\ $$$$\mathrm{2}{p}^{\mathrm{2}} {r}^{\mathrm{2}} −{p}^{\mathrm{4}} −{q}^{\mathrm{4}} −{r}^{\mathrm{4}} +\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$${p}={a}^{\mathrm{1}/\mathrm{2}} \Rightarrow{p}^{\mathrm{2}} ={a}\Rightarrow{p}^{\mathrm{4}} ={a}^{\mathrm{2}} \\ $$$${q}={b}^{\mathrm{1}/\mathrm{2}} \Rightarrow{q}^{\mathrm{2}} ={b}\Rightarrow{q}^{\mathrm{4}} ={b}^{\mathrm{2}} \\ $$$${r}={c}^{\mathrm{1}/\mathrm{2}} \Rightarrow{r}^{\mathrm{2}} ={c}\Rightarrow{r}^{\mathrm{4}} ={c}^{\mathrm{2}} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} {r}^{\mathrm{2}} −{p}^{\mathrm{4}} −{q}^{\mathrm{4}} −{r}^{\mathrm{4}} +\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{ac}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2bc} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{ab}−\mathrm{2}{bc}−\mathrm{2}{ca}\right) \\ $$
Commented by Tawakalitu. last updated on 06/Sep/16
$${wow},\:{thank}\:{you}\:{sir}.\:{God}\:{bless}\:{you} \\ $$