What is x(1-x^2)^3>7(1-x^2)^3 ?

The solution to x(1-x^2)^3>7(1-x^2)^3 is x<-1\lor 1<x<7

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x(1-x^2)^3>7(1-x^2)^3

x (1 - x^{2})^{3} > 7 (1 - x^{2})^{3}

x < - 1 or 1 < x < 7

Interval Notation

(- \infty, - 1) \cup (1, 7)

Solution steps

x (1 - x^{2})^{3} > 7 (1 - x^{2})^{3}

Rewrite in standard form

x - 3 x^{3} + 3 x^{5} - x^{7} + 7 x^{6} - 21 x^{4} + 21 x^{2} - 7 > 0

Factor

x - 3 x^{3} + 3 x^{5} - x^{7} + 7 x^{6} - 21 x^{4} + 21 x^{2} - 7 : - (x + 1)^{3} (x - 1)^{3} (x - 7)

- (x + 1)^{3} (x - 1)^{3} (x - 7) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 1)^{3} (x - 1)^{3} (x - 7)) (- 1) < 0 \cdot (- 1)

Simplify

(x + 1)^{3} (x - 1)^{3} (x - 7) < 0

Identify the intervals

x < - 1 or 1 < x < 7

\frac{x ^{5} y ^{2}}{x ^{4} y ^{0}}

x^{2} - 9 x + 14

64 + 64

2 x^{2} = 3 + x

\frac{x}{3} - 15 = - 7

What is x(1-x^2)^3>7(1-x^2)^3 ?
The solution to x(1-x^2)^3>7(1-x^2)^3 is x<-1\lor 1<x<7