Question

Prove the identities:
(1) $\frac{a^2(x-b)(x-c)}{(a-b)(a-c)}+\frac{b^2(x-c)(x-a)}{(b-c)(b-a)}+\frac{c^2(x-a)(x-b)}{(c-a)(c-b)}=x^2$.

Answer 1

L.H.S $=\frac{a^2(x-b)(x-c)}{(a-b)(a-c)}+\frac{b^2(x-c)(x-a)}{(b-c)(b-a)}+\frac{c^2(x-a)(x-b)}{(c-a)(c-b)}$

$=\frac{a^2(x-b)(x-c)(b-c)}{(a-b)(a-c)(b-c)}+\frac{b^2(x-c)(x-a)(c-a)}{(b-c)(b-a)(c-a)}+\frac{c^2(x-a)(x-b)(a-b)}{(c-a)(c-b)(a-b)}$

$=\frac{[a^2(b-c)+b^2(c-a)+c^2(a-b)]x^2-[a^2(b^2-c^2)+b^2(c^2-a^2)+c^2(a^2-b^2)]x+[a^{2}bc(b-c)+b^{2}ac(c-a)+c^{2}ab(a-b)]}{(a-b)(a-c)(b-c)}$

$=\frac{[a^{2}b-a^{2}c+b^{2}c-b^{2}a+c^{2}a-c^{2}b]x^2-[a^2{b}^2-a^2{c}^2+b^2{c}^2-a^2{b}^2+a^2{c}^2-b^2{c}^2]x+\left[abc(ab-ac+bc-ab+ac-bc)\right]}{(a-b)(a-c)(b-c)}$

$=\frac{(a^{2}b-b^{2}a+b^{2}c-a^{2}c+c^{2}a-c^{2}b)x^2}{(a-b)(a-c)(b-c)}$

$=\frac{(ab(a-b)+c^2(b^2-a^2)+c^2(a-b))x^2}{(a-b)(a-c)(b-c)}$

$=\frac{(a-b)(ab-c(b+a)+c^2)x^2}{(a-b)(a-c)(b-c)}$

$=\frac{(a-b)(ab-bc-ac+c^2)x^2}{(a-b)(a-c)(b-c)}$

$=\frac{(a-b)(b-c)(a-c)x^2}{(a-b)(a-c)(b-c)}$

$=x^2$

$=$ R.H.S