Answer:
[tex]\dfrac{(a-8)(a-6)}{(a-4)^2}[/tex]
Step-by-step explanation:
Given expression:
[tex]\dfrac{a^2-64}{a^2-10a+24} \cdot \dfrac{a^2-12a+36}{a^2+4a-32}[/tex]
Factor the numerator and denominator of both fractions:
[tex]\textsf{Apply the Difference of Two Squares formula} \:\:\:x^2-y^2=(x-y)(x+y):[/tex]
[tex]\begin{aligned} a^2-64 & =a^2+8^2 \\ & =(a-8)(a+8)\end{aligned}[/tex]
[tex]\begin{aligned}a^2-10a+24 & =a^2-4a-6a+24\\& = a(a-4)-6(a-4)\\ & = (a-6)(a-4) \end{aligned}[/tex]
[tex]\begin{aligned}a^2-12a+36 & =a^2-6a-6a+36\\& = a(a-6)-6(a-6)\\ & = (a-6)(a-6) \end{aligned}[/tex]
[tex]\begin{aligned}a^2+4a-32 & =a^2+8a-4a-32\\& = a(a+8)-4(a+8)\\ & = (a-4)(a+8) \end{aligned}[/tex]
Therefore:
[tex]\dfrac{(a-8)(a+8)}{(a-6)(a-4)} \cdot \dfrac{(a-6)(a-6)}{(a-4)(a+8)}[/tex]
[tex]\textsf{Apply the fraction rule}: \quad \dfrac{a}{b} \cdot \dfrac{c}{d}=\dfrac{ac}{bd}[/tex]
[tex]\dfrac{(a-8)(a+8)(a-6)(a-6)}{(a-6)(a-4)(a-4)(a+8)}[/tex]
Cancel the common factors (a + 8) and (a - 6):
[tex]\dfrac{(a-8)(a-6)}{(a-4)(a-4)}[/tex]
Simplify the numerator:
[tex]\dfrac{(a-8)(a-6)}{(a-4)^2}[/tex]
