Find the intervals in which $(x - 3) (x^{2} - 1) > 0$

$(- \infty, - 1)$

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$(- 1, 1)$

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$(1, 3)$

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$(3, \infty)$

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Solution

The correct options are
B
$(- 1, 1)$

D
$(3, \infty)$

$(x - 3) (x^{2} - 1) = (x - 3) (x - 1) (x + 1) L e t P = (x - 3) (x - 1) (x + 1)$
Find the points on the number line where P=0
$P = 0 \Rightarrow x ϵ {- 1, 1, 3}$
Split the number line into 4 using these points.
$(- \infty, - 1), (- 1, 1), (1, 3) & (3, \infty) (- \infty, - 1) : (x - 3) < 0, (x - 1) < 0 & (x + 1) < 0 \Rightarrow P < 0 (- 1, 1) : (x - 3) < 0, (x - 1) < 0 & (x + 1) > 0 \Rightarrow P > 0 (1, 3) : (x - 3) < 0, (x - 1) > 0 & (x + 1) > 0 \Rightarrow P < 0 (3, \infty) : (x - 3) > 0, (x - 1) > 0 & (x + 1) > 0 \Rightarrow P > 0$

$∴ P > 0, w h e n x ϵ (- 1, 1) \cup (3, \infty)$

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