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

Question

Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$

I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all of it and got the answer -2. But it took me 1 hour and I also made many silly mistakes. Is there a quicker way than brute force?

Answer 1

Let $$ f(a,b,c)=\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)} $$Observe $f$ is invariant under permutations: $f(a,b,c)=f(b,c,a)=f(c,a,b)$, etc. Further, observe $f(0,1,2)=f(1,0,-1)=-2$. In other words, $f$ evaluates to $-2$ at $12$ points, and $12$ is greater than the sum of the degrees of the numerator and denominator. Thus $f$ is identically constant (provided no two of $a,b,c$ are equal).

Answer 2

hint

Multiply the first term by $ b-c$ to get in the numerator

$$b^3+c^2b-a^2b-b^2c-c^3+a^2c$$ and the others, by permutation

$$c^3+a^2c-b^2c-c^2a-a^3+b^2a$$ $$a^3+b^2a-c^2a-a^2b-b^3+c^2b$$

The result is $ -2$.

Answer 3

Here is a solution using polynomials. Note that $$\begin{align}S&:=\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)} \\&=(a^2+b^2+c^2)\left(\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(a-b)}\right) \\&\phantom{abc}-2\left(\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-c)(c-a)}+\frac{c^2}{(c-a)(c-b)}\right)\,.\end{align}$$ Consider $$p(x):=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$$ and $$q(x):=a^2\frac{(x-b)(x-c)}{(a-b)(a-c)}+b^2\frac{(x-c)(x-a)}{(b-c)(b-a)}+c^2\frac{(x-a)(x-b)}{(c-a)(c-b)}\,.$$ Observe that $p(x)$ and $q(x)$ are polynomials of degree at most $2$ such that $p(t)=1$ and $q(t)=t^2$ for $t\in\{a,b,c\}$. Therefore, $p(x)=1$ and $q(x)=x^2$ identically. If $[x^k]f(x)$ denotes the coefficient of $x^k$ in a polynomial $f(x)$, then we have $$\begin{align}S&=(a^2+b^2+c^2)\Big([x^2]p(x)\Big)-2\Big([x^2]q(x)\Big)\\&=(a^2+b^2+c^2)\cdot 0-2\cdot 1=-2\,.\end{align}$$

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Alternatively, note that $$S=-\frac{P(a,b,c)}{Q(a,b,c)}\,,$$ where $$P(a,b,c):=(b^2+c^2-a^2)(b-c)+(c^2+a^2-b^2)(c-a)+(a^2+b^2-c^2)(a-b)$$ and $$Q(a,b,c):=(b-c)(c-a)(a-b)\,.$$ Note that when two variables are equal, $P(a,b,c)=0$. Thus, $P(a,b,c)$ is divisible by $Q(a,b,c)$. This shows that $$P(a,b,c)=k\,Q(a,b,c)$$ for some constant $k$. Since $P(-1,0,+1)=4$ and $Q(-1,0,+1)=2$, we conclude that $k=2$, whence $S=-k=-2$.

Answer 4

Here we use $$ab(a-b)+bc(b-c)+ca(c-a)=a^2(b-c)+b^2(c-a)+c^2(a-b)=-(a-b)(b-c)(c-a)$$ In short we cab write $$\sum ab(a-b)=\sum a^2(b-c)=-(a-b)(b-c)(c-a)~~~~(1)$$ Then $$F=\frac{a^2+b^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-a)(b-c)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$$ $$\implies F=\frac{(a^2+b^2-a^2)(c-b)+(c^2+a^2-b^2)(a-c)+(a^2+b^2-c^2)(b-a)}{(a-b)(b-c)(c-a)}$$ upon simplification in the numerator six terms: $(\pm a^3\pm b^3 \pm c^3)$ will cancel each other, we will get $6+6=12$ terms in the numerator as $$F=\frac{-\sum a^2(b-c)-\sum ab(a-b)}{(a-b)(b-c)(c-a)}= 2,$$ on using (1).

Answer 5

First of all, we must have $$a\ne b\ne c$$

Take out the terms containing $b^2$ in the numerator

$$b^2\left(-\dfrac1{(a-b)(c-a)}+\dfrac1{(b-c)(a-b)}-\dfrac1{(c-a)(b-c)}\right)$$

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

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

$$\implies\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$$ $$=\dfrac2{(a-b)(b-c)(c-a)}\cdot\left( b^2(c-a)+c^2(a-b)+a^2(b-c)\right)$$

Now $b^2(c-a)+c^2(a-b)+a^2(b-c)$

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

$=b^2(c-a)+ca(c-a)-b(c^2-a^2)$

$=(c-a)(b^2-b(c+a)+ca)$

$=\cdots$

$=-(a-b)(b-c)(c-a)$