What is x(1-x^2)^3>5(1-x^2)^3 ?

The solution to x(1-x^2)^3>5(1-x^2)^3 is x<-1\lor 1<x<5

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x(1-x^2)^3>5(1-x^2)^3

x (1 - x^{2})^{3} > 5 (1 - x^{2})^{3}

x < - 1 or 1 < x < 5

Interval Notation

(- \infty, - 1) \cup (1, 5)

Solution steps

x (1 - x^{2})^{3} > 5 (1 - x^{2})^{3}

Rewrite in standard form

x - 3 x^{3} + 3 x^{5} - x^{7} + 5 x^{6} - 15 x^{4} + 15 x^{2} - 5 > 0

Factor

x - 3 x^{3} + 3 x^{5} - x^{7} + 5 x^{6} - 15 x^{4} + 15 x^{2} - 5 : - (x + 1)^{3} (x - 1)^{3} (x - 5)

- (x + 1)^{3} (x - 1)^{3} (x - 5) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 1)^{3} (x - 1)^{3} (x - 5)) (- 1) < 0 \cdot (- 1)

Simplify

(x + 1)^{3} (x - 1)^{3} (x - 5) < 0

Identify the intervals

x < - 1 or 1 < x < 5

\frac{3 + 3}{4 12}

∣ 4 x - 6 ∣ = 14

2 x^{2} = x + 1

\frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y}

72 x - 50 = 68 x - 42

What is x(1-x^2)^3>5(1-x^2)^3 ?
The solution to x(1-x^2)^3>5(1-x^2)^3 is x<-1\lor 1<x<5