Posted by **Janice - Please Help!!!** on Tuesday, June 7, 2011 at 4:12pm.

Find the only positive integer whose cube is the sum of the cubes of three positive integers immediately preceding it. Find this positive integer. Your algebraic work must be detailed enough to show this is the only positive integer with this property

- MATH -
**MathMate**, Tuesday, June 7, 2011 at 4:45pm
The property translates to the following equation:

x^3-(x-1)^3-(x-2)^3-(x-3)^3=0

Expanding:

-2x^3+18x^2-42x+36 = 0

2(-x³+9x²-24x+18)=0

which factorizes to:

-2(x-6)(x^2-3x+3)=0

or simply:

(x-6)(x^2-3x+3)=0

The first factor gives x=6 (our answer), and the second factor is not further factorizable.

Attempts to solve

(x^2-3x+3)=0

results in a complex number, so no other real roots exist.

## Answer This Question

## Related Questions

- algebra - Find two consecutive positive integers such that the sum of their ...
- math, algebra - 2a+2ab+2b I need a lot of help in this one. it says find two ...
- math - Find the sum of the first one thousand positive integers. Explain how you...
- math - Prove that a number 10^(3n+1) , where n is a positive integer, cannot be ...
- Maths - Prove that a number 10^(3n+1) , where n is a positive integer, cannot be...
- math - Which statement is true? A.The sum of two positive integers is sometimes ...
- Discrete Math - Let n be positive integer greater than 1. We call n prime if the...
- Math - Tell whether the difference between the two integers is always, sometimes...
- Discrete Math - Theorem: For every integer n, if x and y are positive integers ...
- Math - Paulo withdraws the same amount from his bank account each week to pay ...

More Related Questions