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

Answer:

$\dfrac{(a-8)(a-6)}{(a-4)^2}$

Step-by-step explanation:

                           

Given expression:

$\dfrac{a^2-64}{a^2-10a+24} \cdot \dfrac{a^2-12a+36}{a^2+4a-32}$

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:

$\dfrac{(a-8)(a+8)}{(a-6)(a-4)} \cdot \dfrac{(a-6)(a-6)}{(a-4)(a+8)}$

$\textsf{Apply the fraction rule}: \quad \dfrac{a}{b} \cdot \dfrac{c}{d}=\dfrac{ac}{bd}$

$\dfrac{(a-8)(a+8)(a-6)(a-6)}{(a-6)(a-4)(a-4)(a+8)}$

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

$\dfrac{(a-8)(a-6)}{(a-4)(a-4)}$

Simplify the numerator:

$\dfrac{(a-8)(a-6)}{(a-4)^2}$