Multiply as indicated and simplify: (x^2 - 14x + 48)/(x^2 - 3x - 18) times (x^2 - 9)/(x^2 - 64).

Step-by-step solution

1. 1

   Factor everything before multiplying anything. Expanding the numerators and denominators first would give quartics and hide all the cancellation. Every one of the four quadratics factors over the integers, so factoring is both possible and the whole point.

2. 2

   Factor the four quadratics.

   $x^2-14x+48 = (x-6)(x-8), \qquad x^2-3x-18 = (x-6)(x+3)$

   $x^2-9 = (x-3)(x+3), \qquad x^2-64 = (x-8)(x+8)$

   The last two are differences of squares; the first two need pairs multiplying to $48$ and $-18$ with sums $-14$ and $-3$.

3. 3

   Write the product as one fraction.

   $\frac{(x-6)(x-8)\,(x-3)(x+3)}{(x-6)(x+3)\,(x-8)(x+8)}$

4. 4

   Cancel matched factors, not matched terms. Three whole factors appear top and bottom: $(x-6)$, $(x-8)$ and $(x+3)$. Cancelling them leaves

   $\frac{x-3}{x+8}$

   What you may not do is cancel the bare $x$ or the $3$ in $\dfrac{x-3}{x+8}$ — cancellation only removes identical factors of a product, never pieces of a sum.

5. 5

   Note the excluded values. Cancelling hides restrictions that the original expression still had: the domain excludes $x = 6$, $x = -3$, $x = 8$ and $x = -8$. The simplified form is equal to the original only where the original is defined.

6. 6

   Verify at a test value. Take $x = 1$. Original: $\dfrac{1-14+48}{1-3-18} \cdot \dfrac{1-9}{1-64} = \dfrac{35}{-20} \cdot \dfrac{-8}{-63} = -\dfrac{2}{9}$. Simplified: $\dfrac{1-3}{1+8} = -\dfrac{2}{9}$ ✓.