2021/084) a,b,c are positive real numbers such that $a^2+b^2= c^2 $ and $ab=c$ find the value of $\frac{(a+b+c)(a+b-c)(a-b+c)(c-a+b)}{c^2}$

 We have

$a^2+b^=c^2\cdots(1)$

And $ab = c\cdots(2)$

Now

$\frac{(a+b+c)(a+b-c)}{c} = \frac{(a+b^2) - c^2 }{c} = \frac{a^2+b^2+2ab-c^2}{c} = \frac{2ab}{c}$ (uing (1)

$ = \frac{2c}{c}$ (sing (2)

$ = 2$

Or

$\frac{(a+b+c)(a+b-c)}{c^2} = 2$

similarly

$\frac{(a-b+c)(-a+b+c}{c} = 2$

multiplying we get $\frac{(a+b+c)(a+b-c)(a-b+c)(c-a+b)}{c^2}= 4$