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Solve 33 number please​

Solve 33 number please​-example-1

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Explanation:

let's do the squaring on the right side :

1/(b-c)² + 1/(b-c)(c-a) + 1/(b-c)(a-b) + 1/(b-c)(c-a) + 1/(c-a)² +

+ 1/(c-a)(a-b) + 1/(b-c)(a-b) + 1/(c-a)(a-b) + 1/(a-b)² =

= 1/(b-c)² + 1/(c-a)² + 1/(a-b)² +

+ 2/(b-c)(c-a) + 2/(b-c)(a-b) + 2/(c-a)(a-b)

that means, that in order for the original equation to be true

2/(b-c)(c-a) + 2/(b-c)(a-b) + 2/(c-a)(a-b) = 0

as the first part of the right side is exactly the left side of the original equation.

when

2/(b-c)(c-a) + 2/(b-c)(a-b) + 2/(c-a)(a-b) = 0

then also

1/(b-c)(c-a) + 1/(b-c)(a-b) + 1/(c-a)(a-b) = 0

since we can divide on both sides, we can also multiply and keep the overall equation true.

so, we multiply both sides by e.g. (b-c)(c-a) and get

1/1 + (b-c)(c-a)/(b-c)(a-b) + (b-c)(c-a)/(c-a)(a-b) = 0

1/1 + (c-a)/(a-b) + (b-c)/(a-b) = 0

let's now multiply with (a-b) :

(a-b) + (c-a) + (b-c) = 0

a-a + b-b + c-c = 0

0 = 0

therefore, the original expression is true.

User Bas Van Ommen
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