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Complex (
Posted on 20181222
)
A = 1 + x
^{3}
/3! + x
^{6}
/6! + x
^{9}
/9! + ...
B = x + x
^{4}
/4! + x
^{7}
/7! + x
^{10}
/10! + ...
C = x
^{2}
/2! + x
^{5}
/5! + x
^{8}
/8! + x
^{11}
/11! + ...
Find the value of A
^{3}
+ B
^{3}
+ C
^{3}
 3ABC.
No Solution Yet
Submitted by
Danish Ahmed Khan
Rating:
4.0000
(1 votes)
Comments: (
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)
A start
 Comment 1 of 3
A + B + C = taylor series for e^x,
so A + B + C = e^x
Then e^3x = (A + B + C)^3
expanding and rearranging gives
A^3 + B^3 + C^3  3ABC
= e^3x 3(AB + AC +BC)(A+B+C)
= e^3x 3(AB + AC + BC)e^x
That's as far as I got.
Posted by
Steve Herman
on 20181223 18:10:40
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