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0=(1)/(3)(x-3)(x+3)(x+1)
We move all terms to the left:
0-((1)/(3)(x-3)(x+3)(x+1))=0
Domain of the equation: 3(x-3)(x+3)(x+1))!=0We add all the numbers together, and all the variables
x∈R
-(1/3(x-3)(x+3)(x+1))=0
We multiply parentheses ..
-(1/3(+x^2+3x-3x-9)(x+1))=0
We multiply all the terms by the denominator
-(1=0
We add all the numbers together, and all the variables
=0
x=0/1
x=0
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