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Solve 1x+a+...

Question

Solve $\frac{1}{x + a} + \frac{1}{x + 2 a} + \frac{1}{x + 3 a} = \frac{3}{x}$

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Solution

$\frac{1}{x + a} + \frac{1}{x + 2 a} + \frac{1}{x + 3 a} = \frac{3}{x}$
$(x + 2 a) (x + 3 a) + (x + a) (x + 3 a) + (x + a) (x + 2 a) = \frac{3}{x} (x + a) (x + 2 a) (x + 3 a)$
$x^{2} + 5 a x + 6 x^{2} + x^{2} + 4 a x + 3 a^{2} + x^{2} + 3 x a + 2 a^{2} = \frac{3}{x} [(x^{2} + 3 x a + 2 a^{2}) (x + 3 a)]$
$3 x^{2} + 12 a x + 11 a^{2} = \frac{3}{x} (x^{3} + 3 x^{2} a + 2 x a^{2} + 3 a x^{2} + 9 x a^{2} + 6 a^{3})$
$3 x^{2} + 12 a x^{2} + 11 a^{2} x = 3 x^{2} + 18 x^{3} a + 33 x a^{2} + 18 a^{2}$
$6 x^{2} a + 22 a^{2} x + 18 a^{3} = 0$
$\Rightarrow 3 x^{2} a + 11 a x + 9 a^{2} = 0$
$x = \frac{- 11 a \pm \sqrt{121 a^{2} - 108 a^{2}}}{6} = \frac{- 11 \pm 9 \sqrt{13}}{6}$

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