If 3x(x−6)(x+a)=2x−6+1x+a then 'a'=

3x(x−6)(x+a)=2x−6+1x+a⇒3x(x−6)(x+a)=2(x+a)+1(x−6)(x−6)(x+a)⇒3x=2(x+a)+1(x−6)If x=6⇒18=2(6+a)⇒9=6+a∴a=3.

3x(x−6)(x+a)=2x−6+1x+a⇒3x(x−6)(x+a)=2(x+a)+1(x−6)(x−6)(x+a)⇒3x=2(x+a)+1(x−6)If x=6⇒18=2(6+a)⇒9=6+a∴a=3.

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If $\frac{3 x}{(x - 6) (x + a)} = \frac{2}{x - 6} + \frac{1}{x + a}$ then $' a' =$

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$\frac{3 x}{(x - 6) (x + a)} = \frac{2}{x - 6} + \frac{1}{x + a}$

$\Rightarrow \frac{3 x}{(x - 6) (x + a)} = \frac{2 (x + a) + 1 (x - 6)}{(x - 6) (x + a)}$

$\Rightarrow 3 x = 2 (x + a) + 1 (x - 6)$

If $x = 6 \Rightarrow 18 = 2 (6 + a)$

$\Rightarrow 9 = 6 + a$

$∴ a = 3$ .

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