Find the coefficients of the equivalent polynomial
P ( x , y ) = 5 x 3 + 5 y 3 + 4 x 2 y + 4 x y 2 + 3 x 2 + 3 y 2 + 2 x y + x + y
r = x + y
s = x y
Q ( r , s ) = q 1 r 3 + q 2 r 2 + q 3 r s + q 4 r + q 5 s
If P ( x , y ) = Q ( r , s ) , then q 1 + q 2 + q 3 + q 4 + q 5 = ?
The answer is -6.
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This problem is an application of the fundamental theorem of symmetric polynomials.
By expanding Q ( r , s ) in terms of x and y , we obtain P ( x , y ) = q 1 ( x 3 + y 3 + 3 x 2 y + 3 x y 2 ) + q 2 ( x 2 + y 2 + 2 x y ) + q 3 ( x 2 y + x y 2 ) + q 4 ( x + y ) + q 5 ( x y )
This gives us a system of equations that we can solve for q 1 , q 2 , q 3 , q 4 , and q 5 :
q 1 = 5
q 2 = 3
3 q 1 + q 3 = 4
q 4 = 1
2 q 2 + q 5 = 2
Solving this system equations yields q 3 = − 1 1 and q 5 = − 4 . Thus, q 1 + q 2 + q 3 + q 4 + q 5 = − 6