Show that (x - 1) is a factor of x³ - 7x² + 14x - 8. Hence, completely factorise the given expression.

x - 1 = 0 ⇒ x = 1.

Substituting x = 1 in x³ - 7x² + 14x - 8 we get,

= (1)³ - 7(1)² + 14(1) - 8

= 1 - 7 + 14 - 8

= 15 - 15

= 0.

Since, remainder = 0.

Hence, (x - 1) is a factor of x³ - 7x² + 14x - 8.

On dividing, x³ - 7x² + 14x - 8 by (x - 1),

$\begin{array}{l} \phantom{x - 1)}{x^2 - 6x + 8} \ x - 1\overline{\smash{\big)}x^3 - 7x^2 + 14x - 8} \ \phantom{x - 1}\underline{\underset{-}{}x^3 \underset{+}{-} x^2} \ \phantom{{x - 1}3x^3+}-6x^2 + 14x \ \phantom{{x - 1}3x^3\quad\enspace}\underline{\underset{+}{-}6x^2 \underset{-}{+} 6x} \ \phantom{{x - 1}{3x^3+1}{+2}\qquad}8x - 8 \ \phantom{{x - 1}{2x^3+}{+2x^2}{4}\quad}\underline{\underset{-}{}8x \underset{+}{-} 8} \ \phantom{{x - 1}{2x^3+}{+2x^2-}{4x}\quad}\times \end{array}$

we get, quotient = x² - 6x + 8.

Factorising x² - 6x + 8,

= x² - 4x - 2x + 8

= x(x - 4) - 2(x - 4)

= (x - 2)(x - 4).

Hence, x³ - 7x² + 14x - 8 = (x - 1)(x - 2)(x - 4).