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

`((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(1)`
When x = a,
L.H.S. = `((b-a)(c-a))/((b-a)(c-a))`= 1 R.H.S.
Similarly, when x = -b,
L.H.S. = `((c-b)(a-b))/((c-b)(a-b))`= 1 R.H.S.
When x = -c,
L.H.S. = `((a-c)(b-c))/((a-c)(b-c))`= 1 R.H.S.
Thus, the highest powe of x occurring in relation (1) is e and this relatiom is satisfied bu three distinct values a, b and c of x, therefoue , it is an identity and satisfied by all real values of x.