Interesting answer
If the equation x 3 + 1 1 x 2 − 1 1 + ( x − 1 ) = 1 has the same solutions as the equation x 3 + b x 2 + c x + d . Find b + c + d .
The answer is 0.
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1 solution
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The equation can be simplified as,
( x + 1 ) ( x − 1 ) x ⋅ ( x + 1 ) ( x 2 − x + 1 ) = 1 ⟹ x 3 − x 2 + x = x − 1 ⟹ x 3 − x 2 + 1 = 0
On comparing with the given general equation, we have,
b = ( − 1 ) , c = 0 , d = 1 ⟹ b + c + d = 0
If b = − 1 , c = 0 , d = − 1 then b + c + d = − 2
The equation can be simplified as [x/((x+1)(x-1))] (x+1)(x2 – x +1) = 1. So x3 – x2 + x = x – 1, or x3 – x2 + 1 = 0. b=-1, c=0, d=-1, and b+c+d = 0