Let \(f\left(x\right)=\dfrac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\dfrac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}-1\)

We have:

  \(f\left(a\right)=0+1+0-1=0\)

Similarity \(f\left(b\right)=0,f\left(c\right)=0\)

f(x) is a degree of 2 and have 3 different solotutions (a, b, c) .

=> f(x) = 0

=> \(\dfrac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\dfrac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}=1\)

Let f(x)=(x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)−1f(x)=(x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)−1

We have:

  f(a)=0+1+0−1=0f(a)=0+1+0−1=0

Similarity f(b)=0,f(c)=0f(b)=0,f(c)=0

f(x) is a degree of 2 and have 3 different solotutions (a, b, c) .

=> f(x) = 0

=> (x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)=1

Let f(x)=(x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)−1f(x)=(x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)−1

We have:

  f(a)=0+1+0−1=0f(a)=0+1+0−1=0

Similarity f(b)=0,f(c)=0f(b)=0,f(c)=0

f(x) is a degree of 2 and have 3 different solotutions (a, b, c) .

=> f(x) = 0

=> (x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)=1                                                             Study Well !

Let \(f\left(x\right)=\dfrac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\dfrac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}-1\)

We have:

\(f\left(a\right)=0+1+0-1=0\)

Similarity \(f\left(b\right)=0,f\left(c\right)=0\)

\(f\left(x\right)\) is a degree of 2 and have 3 different solotutions \(\left(a,b,c\right)\)

\(\Rightarrow f\left(x\right)=0\)

\(\Rightarrow\dfrac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\dfrac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}=1\)