Alg5.1.13

Write $\frac{x^3}{(x-a)(x-b)(x-c)}\,$ as partial fractions.

Since the degrees of numerator and denominator are equal in the given expression,let us write

$\frac{x^3}{(x-a)(x-b)(x-c)}=k+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\,$

Equating the numerators on both sides,we get

$x^3=k(x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)\,$

Substituting $x=a\,$

$A=\frac{a^3}{(a-b)(a-c)}\,$

Similarly substituting $x=b,c\,$ we get

$B=\frac{b^3}{(b-a)(b-c)},C=\frac{c^3}{(c-a)(c-b)}\,$

Comparing the coefficients of $x^3\,$ on both sides,we get $k=1\,$

Therefore, $\frac{x^3}{(x-a)(x-b)(x-c)}=1+\frac{a^3}{(a-b)(a-c)(x-a)}+\frac{b^3}{(b-c)(b-a)(x-b)}+\frac{c^3}{(c-a)(c-b)(x-c)}\,$