Question

11a(cube)b(cube)(7c-35)÷3a(square)b(square)(c-5)

Answer 1

we have ,

$\frac{11 {a}^{3} {b}^{3} (7c - 35)}{3 {a}^{2} {b}^{2}(c - 5) }$

Putting 7(c-5) in place of (7c - 35) and dividing a³b³ by a²b², we get,

$\frac{11ab \times 7(c - 5)}{3(c - 5)} \\ = > \frac{77ab}{3}$

Regards

KSHITIJ