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robertcoe426
10.07.2019 •
Mathematics
What are three consecutive numbers whose cubes have a sum of 2241?
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Ответ:
Let "x" be the first integer:
Second integer would be x+1,
Third integer would be x+2,
x³ + (x+1)³ + (x+2)³ = 2241Using the formula;
(a+b)³ = a³ + 3a²b + 3ab² + b³x³ + (x+1)³ + (x+2)³ = 2241
x³ + x³ + 3x² + 3x + 1³ + x³ + 6x² + 12x + 2³ = 2241
3x³ + 9x² + 15x + 9 = 2241
dividing the whole equation by 3;
x³ + 3x² + 5x + 3 = 747
x³ + 3x² + 5x = 747 - 3
x³ + 3x² + 5x = 744By graphing this equation, we can see that there is only 1 real root,
that is:
x = 8
Hence,
x³ = 512(x+1)³ = 729(x+2)³ = 1000Ответ:
Factorial
Step-by-step explanation:
n! gives the no. of arrangements of n distinct objects