b^3-5b^2(-2b+8)/(b-4)

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Solution for b^3-5b^2(-2b+8)/(b-4) equation:


D( b )

b-4 = 0

b-4 = 0

b-4 = 0

b-4 = 0 // + 4

b = 4

b in (-oo:4) U (4:+oo)

b^3-((5*b^2*(8-2*b))/(b-4)) = 0

b^3-5*b^2*(8-2*b)*(b-4)^-1 = 0

(-5*b^2*(8-2*b))/(b-4)+b^3 = 0

(-5*b^2*(8-2*b))/(b-4)+(b^3*(b-4))/(b-4) = 0

b^3*(b-4)-5*b^2*(8-2*b) = 0

b^4+6*b^3-40*b^2 = 0

b^4+6*b^3-40*b^2 = 0

b^2*(b^2+6*b-40) = 0

b^2+6*b-40 = 0

DELTA = 6^2-(-40*1*4)

DELTA = 196

DELTA > 0

b = (196^(1/2)-6)/(1*2) or b = (-196^(1/2)-6)/(1*2)

b = 4 or b = -10

b^2*(b+10)*(b-4) = 0

(b^2*(b+10)*(b-4))/(b-4) = 0

(b^2*(b+10)*(b-4))/(b-4) = 0 // * b-4

b^2*(b+10)*(b-4) = 0

( b+10 )

b+10 = 0 // - 10

b = -10

( b-4 )

b-4 = 0 // + 4

b = 4

( b^2 )

1*b^2 = 0 // : 1

b^2 = 0

b = 0

b in { 4}

b in { -10, 0 }

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