MHB Simplifying Expression: $\frac{z^2(w-x)(x-y)}{(w-x)(x-y)(y-w)}$

[anemone]

Simplify the expression below:

$\dfrac{w^2(z-x)(z-y)}{(w-x)(w-y)}+\dfrac{x^2(z-y)(z-w)}{(x-y)(x-w)}+\dfrac{y^2(z-w)(z-x)}{(y-w)(y-x)}$

 

[anemone]

Hint:

Spoiler

Lagrange's Interpolation Formula

 

[anemone]

Simplify the expression below:

$\dfrac{w^2(z-x)(z-y)}{(w-x)(w-y)}+\dfrac{x^2(z-y)(z-w)}{(x-y)(x-w)}+\dfrac{y^2(z-w)(z-x)}{(y-w)(y-x)}$

Solution of other:

Spoiler

If $f(a)=a^2$ and if we have the points $(a,\,a^2)=(w,\,w^2),\,(x,\,x^2),\,(y,\,y^2)$ then

$g(a)=\dfrac{w^2(a-x)(a-y)}{(w-x)(w-y)}+\dfrac{x^2(a-y)(a-w)}{(x-y)(x-w)}+\dfrac{y^2(a-w)(a-x)}{(y-w)(y-x)}$

We see that $g(a)$ is identical to $f(a)$.

Since $g(a)$ has degree at most two, if we let

$K=\dfrac{w^2(z-x)(z-y)}{(w-x)(w-y)}+\dfrac{x^2(z-y)(z-w)}{(x-y)(x-w)}+\dfrac{y^2(z-w)(z-x)}{(y-w)(y-x)}$, we have

$K=f(z)=z^2$ and we are done.

 

[Albert1]

my solution is almost the same as anemone has mentioned

Spoiler

we consider the original expression as a function of $z$ with degree 2
if $z=x$ then $f(z)=x^2$
if $z=y$ then $f(z)=y^2$
if $z=w$ then $f(z)=w^2$
so we may conclude :$f(z)=z^2$