Question

11a3b3(7c-35)/3a3b3(c-5)​

Answer 1

Answer:

1 Factor out the common term 77

11a\times 3b\times 3\times \frac{7(c-5)}{3}a\times 2b\times 2(c-5)11a×3b×3×​3​​7(c−5)​​a×2b×2(c−5)

2 Regroup terms

\frac{11\times 3\times 3\times 2\times 2}{3}aabb\times 7(c-5)(c-5)​3​​11×3×3×2×2​​aabb×7(c−5)(c−5)

3 Simplify 11\times 3\times 3\times 2\times 211×3×3×2×2 to 396396

\frac{396}{3}aabb\times 7(c-5)(c-5)​3​​396​​aabb×7(c−5)(c−5)

4 Simplify \frac{396}{3}​3​​396​​ to 132132

132aabb\times 7(c-5)(c-5)132aabb×7(c−5)(c−5)

5 Take out the constants

(132\times 7)aabb(c-5)(c-5)(132×7)aabb(c−5)(c−5)

6 Simplify 132\times 7132×7 to 924924

924aabb(c-5)(c-5)924aabb(c−5)(c−5)

7 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x​a​​x​b​​=x​a+b​​

924{a}^{2}{b}^{2}{(c-5)}^{2}924a​2​​b​2​​(c−5)​2

Step-by-step explanation: