Let $f, g, h$ are functions defined by $f (x) = x - 1, g (x) = x^{2} - 2$ and $h (x) = x^{3} - 3$ , show that $(f \circ g) \circ h = f \circ (g \circ h)$ .

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Solution

Given that:
$f (x) = x - 1, g (x) = x^{2} - 2$ and $h (x) = x^{3} - 3$
To show:
$(f \circ g) \circ h = f \circ (g \circ h)$
Solution:
$f \circ g = f [g (x)]$
$= x^{2} - 2 - 1$
$= x^{2} - 3$

$(f \circ g) \circ h = f \circ g [h (x)]$
$= f \circ g (x^{3} - 3)$
$= [(x^{3} - 3)^{2} - 3]$
$= x^{6} - 6 x^{3} + 9 - 3$
$= x^{6} - 6 x^{3} + 6$ $- - - - - - - - - - - - - - - (1)$

$g \circ h = g [h (x)]$
$= g (x^{3} - 3)$
$= (x^{3} - 3)^{2} - 2$
$= x^{6} - 6 x^{3} + 7$

$f \circ (g \circ h) = f (g [h (x)])$
$= f [x^{6} - 6 x^{3} + 7]$
$= x^{6} - 6 x^{3} + 7 - 1$
$= x^{6} - 6 x^{3} + 6$ $- - - - - - - - - - - - - - - (2)$

From eqn. $(1)$ and $(2),$
$(f \circ g) \circ h = f \circ (g \circ h)$

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