Question

6. There are two numbers such that product of the numbers is 45 and their difference is. 12. Find the sum and the difference of their cubes.​

Answer 1

Answer:

yes posible$lhs=rhs x^{2} =per ⬛$

Step-by-step explanation:

Answer 2

Let the numbers be x and y respectively,

$xy = 45(1)$

$x - y = 12(2)$

$let \: x = 12 + y(3)$

$Substitute \: 3 \: in \: 1$

$we \: get \: (12 + y)y = 45 \: we \: get$

$12y + {y}^{2} - 45 = 0$