Answer :

_{Given:}

, where a, b, c are positive real numbers

Here, we have

⇒b(a + b-c) = c(a-b + c)

⇒ab + b^{2}-bc = ac-bc + c^{2}(bc is on both the sides and they get subtracted)

⇒ab + b^{2} = ac + c^{2}

⇒ab-ac = c^{2}-b^{2}

⇒a(b-c) = (c-b)(c + b)

⇒-a(c-b) = (c + b)(c-b)

⇒-a = (c + b)[Here (c-b) is on both the sides,so when they get divided and the result is 1]

⇒(b + c) = -a(**commutative property**)

We have ,

⇒a(a-b + c) = b(-a + b + c)

⇒a^{2}-ab + ac = -ab + b^{2} + bc

⇒a^{2} + ac = b^{2} + bc

⇒a^{2}-b^{2} = bc-ac

⇒(a + b)(a-b) = -c(a-b)

⇒a + b = -c

We have ,

⇒a(a + b-c) = c(-a + b + c)

⇒a^{2} + ab-ac = -ac + bc + c^{2}

⇒a^{2} + ab = bc + c^{2}

⇒a^{2}-c^{2} = bc-ab

⇒(a + c)(a-c) = b(c-a)

⇒(a + c)(a-c) = -b(a-c)

⇒a + c = -b

Therefore, = -1(Here we have substituted the values of (a + b),(b + c),(c + a) from above evaluation)

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