In triangle ABC, with usual notations, (a + b | vedclass

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(A) The given expression is $(a + b + c)(b + c - a)(c + a - b)(a + b - c) = 3b^2c^2$.Using the identity $(a+b+c)(b+c-a) = (b+c)^2 - a^2$ and $(c+a-b)(

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$60^\circ 	ext{ or } 120^\circ$

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(A) The given expression is $(a + b + c)(b + c - a)(c + a - b)(a + b - c) = 3b^2c^2$.Using the identity $(a+b+c)(b+c-a) = (b+c)^2 - a^2$ and $(c+a-b)(a+b-c) = a^2 - (b-c)^2$, the $LHS$ becomes $((b+c)^2 - a^2)(a^2 - (b-c)^2) = 2b^2c^2 + 2c^2a^2 + 2a^2b^2 - a^4 - b^4 - c^4$.This is equal to $16\Delta^2$ where $\Delta$ is the area of the triangle.By Heron's formula, $16\Delta^2 = 16s(s-a)(s-b)(s-c) = 2b^2c^2 + 2c^2a^2 + 2a^2b^2 - a^4 - b^4 - c^4$.Given $16\Delta^2 = 3b^2c^2$, we have $2b^2c^2 + 2c^2a^2 + 2a^2b^2 - a^4 - b^4 - c^4 = 3b^2c^2$.Rearranging, $2c^2a^2 + 2a^2b^2 - a^4 - b^4 - c^4 = b^2c^2$.Using the cosine rule $\cos A = \frac{b^2+c^2-a^2}{2bc}$, we have $a^2 = b^2+c^2 - 2bc \cos A$.Substituting this into the equation leads to $4b^2c^2 \cos^2 A = b^2c^2$, so $\cos^2 A = \frac{1}{4}$.Thus, $\cos A = \pm \frac{1}{2}$, which gives $\angle A = 60^\circ 	ext{ or } 120^\circ$.

In $\triangle ABC$, with usual notations, $(a + b + c)(b + c - a)(c + a - b)(a + b - c) = 3b^2c^2$, then $\angle A = $

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