(w(w-2)(w+4))/8(w+4)=1

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Solution for (w(w-2)(w+4))/8(w+4)=1 equation:



(w(w-2)(w+4))/8(w+4)=1
We move all terms to the left:
(w(w-2)(w+4))/8(w+4)-(1)=0
Domain of the equation: 8(w+4)!=0
w∈R
We multiply parentheses ..
(w(+w^2+4w-2w-8))/8(w+4)-1=0
We multiply all the terms by the denominator
(w(+w^2+4w-2w-8))-1*8(w+4)=0
We calculate terms in parentheses: +(w(+w^2+4w-2w-8)), so:
w(+w^2+4w-2w-8)
We multiply parentheses
w^3+4w^2-2w^2-8w
We do not support ewpression: w^3

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