What is (18x(x^3-2))/((x^3+1)^3)<= 0 ?

The solution to (18x(x^3-2))/((x^3+1)^3)<= 0 is x<-1\lor 0<= x<= \sqrt[3]{2}

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(18x(x^3-2))/((x^3+1)^3)<= 0

\frac{18 x ( x ^{3} - 2 )}{( x ^{3} + 1 ) ^{3}} \leq 0

Interval Notation

Decimal Notation

x < - 1 or 0 \leq x \leq 1.25992 \dots

Solution steps

\frac{18 x ( x ^{3} - 2 )}{( x ^{3} + 1 ) ^{3}} \leq 0

Rewrite in standard form

\frac{x ( x ^{3} - 2 )}{( x ^{3} + 1 ) ^{3}} \leq 0

Factor

Identify the intervals

\frac{- 1}{2} + 1

6 x^{2} + x - 15 = 0

- 2 x^{2} \leq 4

\frac{x - 3}{2} - \frac{x - 4}{3} < 1

3 < 5 - 2 x

What is (18x(x^3-2))/((x^3+1)^3)<= 0 ?
The solution to (18x(x^3-2))/((x^3+1)^3)<= 0 is x<-1\lor 0<= x<= \sqrt[3]{2}