Factorise :
2(ab+cd)−a2−b2+c2+d22(ab + cd) - a^2 - b^2 + c^2 + d^22(ab+cd)−a2−b2+c2+d2
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2(ab+cd)−a2−b2+c2+d2=2ab+2cd−a2−b2+c2+d2=−a2−b2+2ab+c2+d2+2cd=−(a2+b2−2ab)+(c2+d2+2cd)=−(a−b)2+(c+d)2=(c+d)2−(a−b)2=((c+d)−(a−b))((c+d)+(a−b))=(c+d−a+b)(c+d+a−b)2(ab + cd) - a^2 - b^2 + c^2 + d^2\\[1em] = 2ab + 2cd - a^2 - b^2 + c^2 + d^2\\[1em] = - a^2 - b^2 + 2ab + c^2 + d^2 + 2cd\\[1em] = - (a^2 + b^2 - 2ab) + (c^2 + d^2 + 2cd)\\[1em] = - (a - b)^2 + (c + d)^2\\[1em] = (c + d)^2 - (a - b)^2\\[1em] = \Big((c + d) - (a - b)\Big)\Big((c + d) + (a - b)\Big)\\[1em] = (c + d - a + b)(c + d + a - b)2(ab+cd)−a2−b2+c2+d2=2ab+2cd−a2−b2+c2+d2=−a2−b2+2ab+c2+d2+2cd=−(a2+b2−2ab)+(c2+d2+2cd)=−(a−b)2+(c+d)2=(c+d)2−(a−b)2=((c+d)−(a−b))((c+d)+(a−b))=(c+d−a+b)(c+d+a−b)
Hence, 2(ab+cd)−a2−b2+c2+d2=(c+d−a+b)(c+d+a−b)2(ab + cd) - a^2 - b^2 + c^2 + d^2 = (c + d - a + b)(c + d + a - b)2(ab+cd)−a2−b2+c2+d2=(c+d−a+b)(c+d+a−b)
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