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(x1)^3(x2)^3(x3)^3=0
Expanding:
2x^3+18x^242x+36 = 0
2(x³+9x²24x+18)=0
which factorizes to:
2(x6)(x^23x+3)=0
or simply:
(x6)(x^23x+3)=0
The first factor gives x=6 (our answer), and the second factor is not further factorizable.
Attempts to solve
(x^23x+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 ...
 maths  the non decreasing sequence of odd integers {a1, a2, a3, . . .} = {1,3,...
 math, algebra  2a+2ab+2b I need a lot of help in this one. it says find two ...
 Discrete Math  Let n be positive integer greater than 1. We call n prime if the...
 math  Which statement is true? A.The sum of two positive integers is sometimes ...
 math  Prove that a number 10^(3n+1) , where n is a positive integer, cannot be ...
 Math  Tell whether the difference between the two integers is always, sometimes...
 Math  Paulo withdraws the same amount from his bank account each week to pay ...
 math  Find the sum of the first one thousand positive integers. Explain how you...
 Maths  Prove that a number 10^(3n+1) , where n is a positive integer, cannot be...
More Related Questions