Question

1:2:3 and sum of their cubes is4500 find numbers

Answer 1

Answer:

We can use algebra to solve for the numbers given the information provided.

Let x, y, and z be the three numbers. We know that:

x + y + z = 1 + 2 + 3 = 6 (from the first part of the problem)

x^3 + y^3 + z^3 = 4500 (from the second part of the problem)

To find the values of x, y, and z, we can substitute the first equation into the second equation, using the fact that x + y + z = 6:

x^3 + y^3 + z^3 = 4500

x^3 + y^3 + (6-x-y)^3 = 4500

Expanding the cube of (6-x-y) and simplifying the equation

x^3 + y^3 + (216-18x-18y+3xy) = 4500

x^3 + y^3 + 216 - 18x - 18y + 3xy = 4500

x^3 + y^3 + 3xy - 18x - 18y + 216 = 4500

x^3 + y^3 + 3xy = 4524

We have an equation with x, y, x^3, y^3 and xy in it , we can not find the values of x, y, z without more information.