Question

numbers are in ratio 5: 4
diference of their cules 61 find the
numbers​

Answer 1

Step-by-step explanation:

let the numbers be x and y

x : y = 5 : 4

(x² + y² + xy)(x - y) = difference of cubes

(x² + y² + xy)(x - y) = 61

x = 5k

y = 4k

$\bold{( {(5k)}^{2} + {(4k)}^{2} + (5k)(4k))(5k - 4k) = 61} \\ \\ \bold{ 61 {k}^{3} = 61} \\ \\ \bold{k = 1}$

Now

k = 1

x = 5

y = 4

Hope it helps

$\bold{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy)}$