SOLUTION: The product of three consecutive odd integers is ten less than the cube of the smallest odd integer.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: The product of three consecutive odd integers is ten less than the cube of the smallest odd integer.      Log On


   

Question 1001799: The product of three consecutive odd integers is ten less than the cube of the smallest odd integer.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
You can try n, n+2, n+4 as the "odd" integers and see what happens.
n%28n%2B2%29%28n%2B4%29=-10%2Bn%5E3. Solve for n. Does this work?

n%28n%5E2%2B6n%2B8%29=n%5E3-10
n%5E3%2B6n%5E2%2B8n=n%5E3-10
6n%5E2%2B8n=-10
3n%5E2%2B4n%2B5=0
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(3n 1)(n 5) NO
(3n 5)(n 1) NO
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Use formula for quadratic general solution
n=%28-4%2B-+sqrt%2816-4%2A3%2A5%29%29%2F6---NO because discriminant is negative.


Try ensuring that the numbers are ODD.
They should be like 2n+1, 2n+3, 2n+5.
The description:
%282n%2B1%29%282n%2B3%29%282n%2B5%29=%282n%2B1%29%5E3-10
8n%5E3%2B36n%5E2%2B46n%2B15=8n%5E3%2B12n%5E2%2B6n-9
36n%5E2%2B46n%2B15=12n%5E2%2B6n-9
A few more simplifications steps,
3n%5E2%2B5n%2B3=0
NOW, plug values into general solution formula for a quadratic equation and begin simplifying:
n=%28-5%2B-+sqrt%2825-36%29%29%2F6
Notice that the discriminant IS NEGATIVE.
NO SOLUTION