((3x^2-27)(8x^2)6)/(4(9-3x)(x^2+3x))

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Solution for ((3x^2-27)(8x^2)6)/(4(9-3x)(x^2+3x)) equation:


D( x )

4*(9-(3*x))*(x^2+3*x) = 0

4*(9-(3*x))*(x^2+3*x) = 0

4*(9-(3*x))*(x^2+3*x) = 0

4*(9-3*x)*(x^2+3*x) = 0

x^2+3*x = 0

x*(x+3) = 0

x+3 = 0 // - 3

x = -3

x*(x+3) = 0

4*x*(9-3*x)*(x+3) = 0

( 9-3*x )

9-3*x = 0 // - 9

-3*x = -9 // : -3

x = -9/(-3)

x = 3

( x+3 )

x+3 = 0 // - 3

x = -3

( x )

x = 0

x in (-oo:-3) U (-3:0) U (0:3) U (3:+oo)

((8*x^2)^6*(3*x^2-27))/(4*(9-(3*x))*(x^2+3*x)) = 0

((8*x^2)^6*(3*x^2-27))/(4*(9-3*x)*(x^2+3*x)) = 0

(262144*x^12*(3*x^2-27))/(4*(9-3*x)*(x^2+3*x)) = 0

x^2+3*x = 0

x*(x+3) = 0

x+3 = 0 // - 3

x = -3

x*(x+3) = 0

(262144*x^12*(3*x^2-27))/(4*x*(9-3*x)*(x+3)) = 0

( 262144*x^12 )

262144*x^12 = 0 // : 262144

x^12 = 0

x = 0

( 3*x^2-27 )

3*x^2 = 27 // : 3

x^2 = 9

x^2 = 9 // ^ 1/2

abs(x) = 3

x = 3 or x = -3

x in { 0}

x in { 3}

x in { -3}

x belongs to the empty set

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