Let f(x) = x^3 - 2x + 2. If real numbers a, b | vedclass

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(C) Given $|f(a)| + |f(b)| + |f(c)| = 0$. Since the absolute value is always non-negative,this implies $f(a) = 0, f(b) = 0$,and $f(c) = 0$.Thus,$a, b,

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$36$

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(C) Given $|f(a)| + |f(b)| + |f(c)| = 0$. Since the absolute value is always non-negative,this implies $f(a) = 0, f(b) = 0$,and $f(c) = 0$.Thus,$a, b, c$ are the roots of the cubic equation $x^3 - 2x + 2 = 0$.For any root $x$,we have $x^3 - 2x + 2 = 0$,which implies $x^3 = 2x - 2$.Dividing by $x$ (assuming $x 
eq 0$,which is true as $f(0) = 2 
eq 0$),we get $x^2 = 2 - \frac{2}{x}$,or $x^2 + \frac{2}{x} = 2$.Since $a, b, c$ are roots,$a^2 + \frac{2}{a} = 2$,$b^2 + \frac{2}{b} = 2$,and $c^2 + \frac{2}{c} = 2$.We need to evaluate $f^2(a^2 + \frac{2}{a}) + f^2(b^2 + \frac{2}{b}) - f^2(c^2 + \frac{2}{c}) = f^2(2) + f^2(2) - f^2(2) = f^2(2)$.Calculating $f(2) = 2^3 - 2(2) + 2 = 8 - 4 + 2 = 6$.Therefore,$f^2(2) = 6^2 = 36$.

Let $f(x) = x^3 - 2x + 2$. If real numbers $a, b, c$ are such that $|f(a)| + |f(b)| + |f(c)| = 0$,then the value of $f^2(a^2 + \frac{2}{a}) + f^2(b^2 + \frac{2}{b}) - f^2(c^2 + \frac{2}{c})$ is equal to:

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