Solve using suitable identity:
$x^{4} - 1$ $= 0$

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Solution

$x^{4} - 1 = 0 \Rightarrow (x^{2} - 1) (x^{2} + 1) = 0 \Rightarrow x^{2} - 1 = 0 \Rightarrow (x - 1) (x + 1) = 0 \Rightarrow x = 1, - 1$
[ Since $x$ is real number then $x^{2} + 1 = 0$ does not give any real root]

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