2015/073) Show that the following is true for all real values of x.

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

Solution 2015/073)

This can be done in 2 ways

method 1

expand

we have 1^st term

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

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

= $\dfrac{x^2(- a^2(b-c)) +xa^2(b^2-c^2) -a^2bc(b-c)}{(a-b)(b-c)(c-a)}$

similarly we have 2^nd term

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

= $\dfrac{x^2(- b^2(c-a)) +xb^2(c^2-a^2) -b^2ca(c-a)}{(a-b)(b-c)(c-a)}$

3^rd term

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

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

so the sum = 
$\frac{(x^2(- a^2(b-c)-b^2(c-a) -c^2(a-b)) +x(a^2(b^2-c^2) + b^2(c^2-a^2) + c^2(a^2-b^2)) – (a^2bc(b-c) + b^2ca(c-a) + c^2ab(a^2-b^2)}{(a-b)(b-c)(c-a))}$

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

numerator = $(x^2(- a^2(b-c)-b^2(c-a) -c^2(a-b))$

= $x^2(-a^2(b-c) – b^2c+ ab^2 -ac^2 +bc^2)$

= $x^2(-a^2(b-c) – b^2c+ bc^2 -ac^2 +ab^2)$

= $x^2(-a^2(b-c) – bc(b-c) + a(b^2-c^2))$

= $x^2(-a^2(b-c) – bc(b-c) + a(b+c)(b-c))$

= $x^2(b-c)(-a^2-bc+ab+ac)$

= $x^2(b-c)(a-b)(c-a)$

so we have ratio = $x^2$

hence proved

Aliter (Method 2)

This can be proved as below

because this is quadratic this cannot have more than 2 root.

But x= a, x= b, x= c satisfy the condition.

Hence this has to be identity.

Proved