If $\{x}/{x^2 - 2x + 1} = \{1}/{3}$, then find the value of $x^3 + \{1}/{x^3}$.

Question

Accepted Answer

Step 1: Simplify the denominator:
$x^2 - 2x + 1 = (x-1)^2$.

Step 2: Given equation:
$$ \frac{x}{(x-1)^2} = \frac{1}{3} $$
Cross multiply:
$$ 3x = (x-1)^2 $$
$$ 3x = x^2 - 2x + 1 $$

Step 3: Rearrange to form a quadratic:
$$ x^2 - 2x + 1 - 3x = 0 $$
$$ x^2 - 5x + 1 = 0 $$

Step 4: Let $x + \frac{1}{x} = t$.
From the quadratic, multiply both sides by $\frac{1}{x}$ (assuming $x \neq 0$):
$$ x - 5 + \frac{1}{x} = 0 $$
$$ x + \frac{1}{x} = 5 $$
So, $t = 5$.

Step 5: Find $x^3 + \frac{1}{x^3}$ using the identity:
$$ x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) $$
$$ = t^3 - 3t $$
$$ = 5^3 - 3 \times 5 = 125 - 15 = 110 $$

Step 6: Check the options. 110 is option 1, but the correct calculation shows 110, so option 1 is correct.

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