1÷(x-1)(x-2)+1÷(x-2)(x-3)+1÷(x-3)(x-4)=1/6 solve the value of x

We have,
{tex}\frac{1}{{(x - 1)(x - 2)}} + \frac{1}{{(x - 2)(x - 3)}}{/tex} {tex}+ \frac{1}{{(x - 3)(x - 4)}} = \frac{1}{6}{/tex}
{tex}\Rightarrow{/tex}{tex} (x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2) = {/tex}{tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
[{tex}\because{/tex} Multiplying both sides by (x -1)(x - 2)(x - 3)(x - 4)]
{tex}\Rightarrow{/tex} {tex}x^2 - 4x - 3x + 12 + x^2 - 4x - x + 4 + x^2 - 2x - x + 2 ={/tex} {tex}\frac{1}{6}{/tex}{tex}[(x - 1)(x - 2)(x - 3)(x - 4)]{/tex}
{tex}\Rightarrow{/tex} {tex}3x^2 - 15x + 18 ={/tex} {tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}3(x^2 - 5x + 6) =​​​​​​​{/tex} {tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18[x^2 - 3x - 2x + 6] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18[x(x - 3) - 2(x - 3)] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18(x - 3)(x - 2) = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} 18 = (x - 1)(x - 4)
{tex}\Rightarrow{/tex} 18 = x² - 4x - 1x + 4
{tex}\Rightarrow{/tex} x² - 5x + 4 - 18 = 0
{tex}\Rightarrow{/tex} x² - 5x - 14 = 0
In order to factorize x² - 5x - 14, we have to find two numbers 'a' and 'b' such that.
a + b = - 5 and ab = -14
Clearly, -7 + 2 = -5 and (-7)(2) = -14
{tex}\therefore{/tex} a = -7 and b = 2
Now,
x² - 5x - 14 = 0
{tex}\Rightarrow{/tex} x² - 7x + 2x - 14 = 0
{tex}\Rightarrow{/tex} x(x - 7) + 2(x - 7) = 0
{tex}\Rightarrow{/tex} (x - 7)(x + 2) = 0
{tex}\Rightarrow{/tex} x - 7 = 0 or x + 2 = 0
{tex}\Rightarrow{/tex} x = 7 or x = -2