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
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