3(x^4 + 5x^3 + 8x^2 + 7x + 3) = (x+1)(x+3)(x^2+x+1).
let (x^4 + 5x^3 + 8x^2 + 7x + 3) be t.
then (x^4 + 5x^3 + 8x^2 + 7x + 5) is (t+2).
t^4 + (t+2)^4 = 16
so t = 0 or t = -2.
the other two solutions are imaginary.
Solve the equation (x^4 + 5x^3 + 8x^2 + 7x + 5)^4 + (x^4 + 5x^3 + 8x^2 + 7x + 3)^4 = 16 in the set of real numbers.
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