Find the value of $a$ and $b$ if $\left(x - 1\right)$ and $\left(x - 2\right)$ are factors of $x^3 - ax + b$.
Let $f \left(x\right) = x^3 - ax + b$
$\because \;$ $\left(x - 1\right)$ is a factor of $f \left(x\right)$ $\implies$ by factor theorem $\;$ $f \left(1\right)
= 0$
i.e. $\;$ $f \left(1\right) = 1^3 - a \times 1 + b = 0$
i.e. $\;$ $a - b = 1$ $\;\;\; \cdots \; (1)$
$\because \;$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$ $\implies$ by factor theorem $\;$ $f \left(2\right)
= 0$
i.e. $\;$ $f \left(2\right) = 2^3 - a \times 2 + b = 0$
i.e. $\;$ $2a - b = 8$ $\;\;\; \cdots \; (2)$
Solving equations $(1)$ and $(2)$ simultaneously we have,
$a = 7$ $\;$ and $\;$ $b = 6$