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I am using this method in which the partial fraction is broken in to 3 parts namely A,B &C

$\frac{x^2}{(x-2)(x+3)(x-1)}=\frac{A}{(x-2)}+\frac{B}{(x+3)}+\frac{C}{(x-1)}$

$\frac{x^2}{(x-2)(x+3)(x-1)}=\frac{A(x+3)(x-1)+B(x-2)(x-1)+C(x-2)(x+3)}{(x-2)(x-3)(x-1)}$

Common denominators cancel out resulting in,

${x^2}={A(x+3)(x-1)+B(x-2)(x-1)+C(x-2)(x+3)}$

Now what values should be substituted to $x$ in order to get the relevant values for A,B & C