Question

AHHHHHHHHHHHH this is so hard

Answer 1

`{ \qquad\qquad\huge\underline{{\sf Answer}}}`

Let's solve ~

`\qquad \sf  \dashrightarrow \: \cfrac{ {a}^(2) - 64 }{ {a}^(2) - 10a + 24} \sdot \cfrac{ {a}^(2) - 12a + 36 }{ {a}^(2) + 4a - 32}`

`\qquad \sf  \dashrightarrow \: \cfrac{( {a}^{} + 8)(a - 8) }{ {a}^(2) - 6a - 4a+ 24} \sdot \cfrac{ {a}^(2) - 6a - 6a + 36 }{ {a}^(2) + 8a - 4a- 32}`

`\qquad \sf  \dashrightarrow \: \cfrac{( {a}^{} + 8)(a - 8) }{ {a}^{} (a - 6) - 4(a - 6)} \sdot \cfrac{ {a}^{} (a - 6) - 6(a - 6) }{ {a}^{}(a + 8) - 4(a + 8)}`

`\qquad \sf  \dashrightarrow \: \cfrac{( {a}^{} + 8)(a - 8) }{(a - 6) (a -4)} \sdot \cfrac{ (a - 6) (a - 6) }{ {}^{}(a - 4)(a + 8)}`

`\qquad \sf  \dashrightarrow \: \cfrac{(a - 8) }{(a -4)} \sdot \cfrac{ (a - 6) }{ {}^{}(a - 4)}`

`\qquad \sf  \dashrightarrow \: \cfrac{(a - 8)(a - 6) }{(a -4) {}^(2) }`

Or [ in expanded form ]

`\qquad \sf  \dashrightarrow \: \cfrac{ {a}^(2) - 8a - 6a + 48 }{ {a}^(2) - 8a + 16 }`

`\qquad \sf  \dashrightarrow \: \cfrac{ {a}^(2) -14a + 48 }{ {a}^(2) - 8a + 16 }`

Answer 2

`((a-8)(a-6))/((a-4)^2)`

Explanation:

Given expression:

`(a^2-64)/(a^2-10a+24) \cdot (a^2-12a+36)/(a^2+4a-32)`

Factor the numerator and denominator of both fractions:

`\textsf{Apply the Difference of Two Squares formula} \:\:\:x^2-y^2=(x-y)(x+y):`

`\begin{aligned} a^2-64 & =a^2+8^2 \\ & =(a-8)(a+8)\end{aligned}`

`\begin{aligned}a^2-10a+24 & =a^2-4a-6a+24\\& = a(a-4)-6(a-4)\\ & = (a-6)(a-4) \end{aligned}`

`\begin{aligned}a^2-12a+36 & =a^2-6a-6a+36\\& = a(a-6)-6(a-6)\\ & = (a-6)(a-6) \end{aligned}`

`\begin{aligned}a^2+4a-32 & =a^2+8a-4a-32\\& = a(a+8)-4(a+8)\\ & = (a-4)(a+8) \end{aligned}`

Therefore:

`((a-8)(a+8))/((a-6)(a-4)) \cdot ((a-6)(a-6))/((a-4)(a+8))`

`\textsf{Apply the fraction rule}: \quad (a)/(b) \cdot (c)/(d)=(ac)/(bd)`

`((a-8)(a+8)(a-6)(a-6))/((a-6)(a-4)(a-4)(a+8))`

Cancel the common factors (a + 8) and (a - 6):

`((a-8)(a-6))/((a-4)(a-4))`

Simplify the numerator:

`((a-8)(a-6))/((a-4)^2)`