Question 18
The sum of all the roots of the equation $$\sqrt[3]{16-x^3} = 4-x$$ is ------.
Correct Answer: 4
Solution
Cubing both sides gives $$16-x^3 = (4-x)^3$$, and expanding causes the $$x^3$$ terms to cancel, leaving $$x^2-4x+4=0$$, i.e. $$(x-2)^2=0$$. By Vieta's formulas the sum of the (repeated) roots of this quadratic is $$4$$.
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