algebra question

Daunting.

 

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

From the top line,

\(\displaystyle \frac{a^{2}}{(x-a)}+a=\frac{a^{2}+a(x-a)}{(x-a)}=\frac{ax}{(x-a)}.\)

 

Similarly,

\(\displaystyle \frac{b^{2}}{(x-b)}+b=\frac{bx}{(x-b)}\quad \text{and} \quad \frac{c^{2}}{(x-c)}+c=\frac{cx}{(x-c)}.\)

 

Now put those together, remove the x as a common factor and then cancel.

 

Result is simply x.