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November 29, 2014

November 29, 2014

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

- MATH -
**MathMate**, Tuesday, June 7, 2011 at 4:45pmThe 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.

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