Show that `((x+b)(x+c))/((b-a)(c-a))+((x+c)(x+a))/((c-b)(a-b))+((x+a)(x+b))/((a-c)(b-c))=1` is an identity.

Explanation

Given relation is
`((x+)(x+c))/((b-a)(c-a))+((x+c)(x+a))/((c-b)(a-b))+((x+a)(x+b))/((a-c)(b-c))=1`….i
When `x=-a`, then LHS of eq. (i) `((b-a)(c-a))/((b-a)(c-a))=1`
`=` RHS of Eq. (i)
When `x=` then LHS of eq. (i)
`=((c-b)(a-b))/((c-b)(a-b))=1=` RHS of Eq. (i)
And when `x=-c` then LHS of eq. (i) `=((a-c)(b-c))/((a-c)(b-c))=1`
`=RHS` of eq. (i)
Thus, highest power of `x` occuring in relation of Eq. (i) is 2 and this relation is satisfied by three distinct values o `x(=-a,-b,-c)`.Therefore it cannot be an equation and hence it is an identity.