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26/01/2018 at 07:45
Faded 26/01/2018 at 12:25
Put A(x)=(3−4x+x2)2016.(3+4x+x2)2017
The sum of all coefficients after expanding the expression is:
A(1)=(3−4.1+12)2016.(3+4.1+12)2017
=02016.82017=0

Alone 26/01/2018 at 12:14
Put \(A\left(x\right)=\left(34x+x^2\right)^{2016}.\left(3+4x+x^2\right)^{2017}\)
The sum of all coefficients after expanding the expression is:
\(A\left(1\right)=\left(34.1+1^2\right)^{2016}.\left(3+4.1+1^2\right)^{2017}\)
\(=0^{2016}.8^{2017}=0\)
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