AN ALGEBRA QQQ.

I will derived my own identity.

$(x+y)^2-2xy=x^2+2xy+y^2-2xy=x^2+y^2$ .

Now just plug in the given quantities in the derived formula. We have

$17^2-2xy=167$

$2xy=17^2-167=122$

Finally,

$\color{#D61F06}\large \boxed{xy=61}$

$\qquad As\\ \qquad \qquad x+y=17\quad ................\quad (i)\\ \qquad \quad \quad { x }^{ 2 }+{ y }^{ 2 }=167\quad ............\quad (ii)\\ Taking\quad Squre\quad of\quad (i),\quad we\quad have\\ =>\qquad ({ x }^{ 2 }+{ y }^{ 2 })+2xy=289\\ =>\quad \quad \quad \quad (167)+2xy=289\quad \quad \quad \quad \because { x }^{ 2 }+{ y }^{ 2 }=167\\ =>\quad \quad \quad \quad 2xy=289-167\\ =>\quad \quad \quad \quad 2xy=122\\ =>\quad \quad \quad \quad \quad xy=\frac { 122 }{ 2 } \\ So,\quad \boxed { xy=61 }$

x+y=17 x^2+y^2+2xy=289 2xy=289-167 xy=61