Question

1/(1-x)(1-2x)(1-3x)=
​

Answer 1

Step-by-step explanation:

take the denominator on left side

so, we said that,

1=(1-2x-x+2x²)(1-3x)=(1-3x+2x²)(1-3x)

1=(1-3x-3x+9x²+2x²-6x³)=(1-6x+11x²-6x³)

1=1-x(6+11x-6x²)

so by above form of equation,we say that

-x(6+11x-6x²)=0

now we common the minus inside brackets

-{-x(6x²-11x-6)}=0

x(6x²-11x-6)=0

so we say that

x=0 or 6x²-11x-6=0

but here we don't know the value x=0 is correct or not. So we take other equation

6x²-11x-6=0

now we use Sri dharacharya method

x={-b±√(b²-4ac)}/2a

x={-11±√(11²-4*6*6)}/2*6

x={-11+√121-144}/12 or x={-11-√121-144}/12

x=(-11+√-23)/12 or x={-11-√-23}/12

these are your answers. If now about Sri Dharacharya method then you know it