Question

Two positive numbers x and y are such that x > y if the difference of these numbers is 5 and their products is 24 find 1) sum of these numbers 2) difference of their cubes 3) sum of their cubes

Answer 1

x-y=5,xy=24
x=5+y
put in xy=24
(5+y)y=24
5y+y square =24
y square+5y-24=0
by factorisation method
y=3,-8
we will take y=3 ,then x=5+3=8
so sum of these no.'s is 8+3=11,
difference of their cube is 485,
sum of their cube is 539

Answer 2

Given,

Two numbers x, y.

The difference between that numbers is five.

And the product is 24.

To Find,

The numbers and their sum, their cubes difference, and the sum of their cubes.

Solution,

Given that,

There are two numbers.

Let's take those numbers are x and y.

Then given that the number difference is,

X-Y =5

And the numbers multiplication result is,

XY =24.

To find,
X and Y$8^{3} + 3^{3}$

X = 5+y

(5+Y)×Y =24

5Y + $Y^{2}$ = 24

$Y^{2}$ +5Y-24=0

When do factorisation we will get,

Y = 3, -8.

Then the value of Y = 3.

Therefore,

X = 5+3

X = 8.

(i) To find the sum of the numbers

X+Y = 8+3

X+Y = 11.

(ii) The difference between the cubes

$X^{3} - Y^{3}$ =

485.

(iii) To find the number's cubes sum.

$X^{3} + Y^{3}$ = $8^{3} +3^{3}$

539.

Hence, the numbers are 8 & 3, the sum of the numbers is 11, and their cubes difference is 485, cubes sum is 539.