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Question

Let a,b,c such that $fraction numerator 1 over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction equals fraction numerator a over denominator 1 minus x end fraction plus fraction numerator b over denominator 1 minus 2 x end fraction plus fraction numerator c over denominator 1 minus 3 x end fraction$ then , $fraction numerator a over denominator 1 end fraction plus fraction numerator b over denominator 3 end fraction plus fraction numerator c over denominator 5 end fraction equals$ ?

$\frac{1}{15}$
$\frac{1}{6}$
$\frac{1}{5}$
$\frac{1}{3}$

hint Hint:

Take LCM on RHS, then equate the coefficients by comparing LHS and RHS and find the value of a,b, c

The correct answer is: $fraction numerator 1 over denominator 15 end fraction$

Given :
$fraction numerator 1 over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction equals fraction numerator a over denominator 1 minus x end fraction plus fraction numerator b over denominator 1 minus 2 x end fraction plus fraction numerator c over denominator 1 minus 3 x end fraction$

Taking LCM on RHS
$rightwards double arrow fraction numerator 1 over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction equals space fraction numerator a left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis space plus space b left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 3 x right parenthesis space plus space c left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction$

Simplifying
$rightwards double arrow fraction numerator 1 over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction equals space fraction numerator a left parenthesis 6 x squared space minus space 5 x space plus 1 right parenthesis space plus space b left parenthesis 3 x squared space minus space 4 x space plus 1 right parenthesis space plus space c left parenthesis 2 x squared space minus 3 x space plus 1 right parenthesis over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction$

Cancelling denominators on both side
$rightwards double arrow 1 over blank equals space fraction numerator a left parenthesis 6 x squared space minus space 5 x space plus 1 right parenthesis space plus space b left parenthesis 3 x squared space minus space 4 x space plus 1 right parenthesis space plus space c left parenthesis 2 x squared space minus 3 x space plus 1 right parenthesis over denominator blank end fraction$

Further simplifying
$rightwards double arrow 1 over blank equals space fraction numerator x squared left parenthesis 6 a space plus 3 b space plus 2 c right parenthesis space plus space x left parenthesis negative 5 a space minus space 4 b space minus 3 c right parenthesis space plus space a space plus space b space plus c over denominator blank end fraction$

Equating coefficients by comparing LHS and RHS
$6 a space plus space 3 b space plus space 2 c space equals space 0 space............. left parenthesis 1 right parenthesis minus 5 a space minus space 4 b space minus space 3 c space equals space 0 space rightwards double arrow 5 a space plus space 4 b space plus space 3 c space equals space 0 space...... space left parenthesis 2 right parenthesis a space plus space b space plus space c space equals space 1 space.................. left parenthesis 3 right parenthesis$

Multiplying by 5 in equation 3 and subtracting equation 2 from it, we get
$space space 5 a space plus space 4 b space plus space 3 c space equals space 0 space minus space space 5 a space plus 5 space b space plus 5 space c space space equals 5 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ space space space space space space space space space minus b space minus 2 c space equals space minus 5 space rightwards double arrow space b space plus space 2 c space equals space 5 space........ space left parenthesis 4 right parenthesis$
Multiplying by 6 in equation 3 and subtracting equation 1 from it, we get
$space space space space 6 a space plus space 3 b space plus space 2 c space equals space 0 minus space space space space 6 a space plus space 6 b space plus space 6 c space equals space 6 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ space space space space space space space space space space space minus 3 b space minus 4 c space equals space minus 6 space rightwards double arrow space 3 b space plus space 4 c space equals space 6 space...... space left parenthesis 5 right parenthesis$

Multiplying by 2 in equation 4 and subtracting equation 5 from it, we get
Multiplying by 6 in equation 3 and subtracting equation 1 from it, we get

2b + 4c = 10
-
3b + 4c = 6
______________
- b = 4
b = -4
Substituting value of b, we get c = 9/2 and a = 1/2
$a over 1 space plus space b over 3 space plus space c over 5 space equals space 1 half space plus space fraction numerator negative 4 over denominator 3 end fraction space plus space 9 over 10 space equals space 2 over 30 space equals space 1 over 15$

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