If( a + b) 2+(b+c) 2+(c+d)2=4(ab +bc+cd)
which is correct
1)a+b+c+d=0
2)a=b=c=d
3)a/b+b/c+c/d=1
4)a 2/b 2 +c 2/d 2=1

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Solution

dear student, (a+b)²+(b+c)²+(c+d)²=4(ab+bc+cd) a²+2ab+b²+b²+2bc+c²+c²+2cd+d² = 4ab+4bc+4cd Get 0 on the right by subtracting it from both sides: a²-2ab+b²+b²-2bc+c²+c²-2cd+d² = 0 (a-b)²+(b-c)²+(c-d)² = 0 The three terms on the left are non-negative. Thus they must all be 0. Thus a-b=0, which means a=b. b-c=0, which means b=c, c-d=0, which means c=d So a=b=c=d option 2 is correct

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