Show that the value of $x^{3} - 8 x^{2} + 12 x - 5$ is zero when $x = 1$ .

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Solution

Substitute $x = 1$ in $x^{3} - 8 x^{2} + 12 x - 5$

$x^{3} - 8 x^{2} + 12 x - 5 = (1)^{3} - 8 (1)^{2} + 12 (1) - 5$

$= 1 - 8 (1) + 12 - 5$

$= 1 - 8 + 12 - 5$

$= (1 + 12) - (8 + 5)$

$= 13 - 13$

$= 0$

Therefore, the value of $x^{3} - 8 x^{2} + 12 x - 5$ is zero when $x = 1$ .

Hence proved.

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