# Question #71d2a

Jun 17, 2017

See a solution process below:

#### Explanation:

First, rewrite the third term as:

${x}^{2} \left(a - b\right) - 6 x \left(a - b\right) - 9 \left(- 1 \left(a - b\right)\right) \implies$

${x}^{2} \left(a - b\right) - 6 x \left(a - b\right) + 9 \left(a - b\right)$

We can now factor and $\left(a - b\right)$ out of each term giving:

$\left(a - b\right) \left({x}^{2} - 6 x + 9\right)$

We can now factor the quadratic as:

$\left(a - b\right) \left(x - 3\right) \left(x - 3\right)$

Or

$\left(a - b\right) {\left(x - 3\right)}^{2}$

Jun 17, 2017

${x}^{2} \left(a - b\right) - 6 x \left(a - b\right) - 9 \left(b - a\right)$ can be written as

${x}^{2} \left(a - b\right) - 6 x \left(a - b\right) - \left(- 9\right) \left(a - b\right)$, taking -1 common from the last bracket.
=${x}^{2} \left(a - b\right) - 6 x \left(a - b\right) + 9 \left(a + b\right)$

Taking (a-b) common from the whole expression we get

$\left(a - b\right) \left({x}^{2} - 6 x + 9\right)$

$\left({x}^{2} - 6 x + 9\right)$ is simply ${\left(x - 3\right)}^{2}$

So, the expression can be factorised as

${x}^{2} \left(a - b\right) - 6 x \left(a - b\right) - 9 \left(b - a\right) = \left(a - b\right) {\left(x - 3\right)}^{2}$

Hope this helps.

Feel free to comment if you want further explanation of any step.