Question

• If a^2+b^2 + c ^2= (a+b+c)^2, then ab+bc+ca=​

Answer 1

Step-by-step explanation:

a2+b2+c2-ab-bc-ca=0

Multiply both sides with 2, we get

2( a2 + b2 + c2 – ab – bc – ca) = 0

2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0

(a2 – 2ab + b2) + (b2 – 2bc + c2) + (c2 – 2ca + a2) = 0

(a –b)2 + (b – c)2 + (c – a)2 = 0

Since the sum of square is zero then each term should be zero

(a –b)2 = 0, (b – c)2 = 0, (c – a)2 = 0

(a –b) = 0, (b – c) = 0, (c – a) = 0

a = b, b = c, c = a

a = b = c.

therefore c+a/b=2

Answer 2

Answer:

ab + bc + ca = 0

Step-by-step explanation:

$(x + y + z)^2$ = $x^2 + y^2 + z^2 + 2xy + 2yz + 2xz$  ------------------> 1

So , if $a^2 + b^2 + c^2 = (a + b + c)^2$ , then

From 1,

2ab + 2bc + 2ca = 0

ab + bc + ca = 0

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