If x satisfies the equation x 2 - 2x + 1 = 0, then the value of \(\rm x^3 - \frac{1}{x^3}\) is:
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Solving the Equation \(x^2 - 2x + 1 = 0\)
The question asks us to find the value of the expression \(x^3 - \frac{1}{x^3}\), given that \(x\) satisfies the equation \(x^2 - 2x + 1 = 0\). First, we need to solve the given quadratic equation for \(x\).
The equation \(x^2 - 2x + 1 = 0\) is a perfect square trinomial. It can be factored easily.
The general form of a perfect square trinomial is \(a^2 - 2ab + b^2 = (a-b)^2\). In our equation, \(a^2 = x^2\), which means \(a=x\), and \(b^2 = 1\), which means \(b=1\). The middle term is \(-2x\), which matches \(-2ab = -2(x)(1)\).
So, we can rewrite the equation as:
\((x-1)^2 = 0\)
To find the value of \(x\), we take the square root of both sides of the equation:
\(\sqrt{(x-1)^2} = \sqrt{0}\)
\(x-1 = 0\)
Now, we solve for \(x\):
\(x = 1\)
So, the value of \(x\) that satisfies the given equation \(x^2 - 2x + 1 = 0\) is \(1\).
Evaluating the Expression \(x^3 - \frac{1}{x^3}\)
Now that we have the value of \(x\), which is \(1\), we can substitute this value into the expression \(x^3 - \frac{1}{x^3}\).
Substitute \(x=1\) into the expression:
\(1^3 - \frac{1}{1^3}\)
Calculate the powers of 1:
- \(1^3 = 1 \times 1 \times 1 = 1\)
- \(\frac{1}{1^3} = \frac{1}{1} = 1\)
Now substitute these values back into the expression:
\(1 - 1\)
Perform the subtraction:
\(1 - 1 = 0\)
Therefore, the value of \(x^3 - \frac{1}{x^3}\) when \(x\) satisfies \(x^2 - 2x + 1 = 0\) is \(0\).
Step-by-Step Solution Summary
- Solve the equation \(x^2 - 2x + 1 = 0\).
- Recognize the equation as a perfect square trinomial: \((x-1)^2 = 0\).
- Solve for \(x\): \(x-1 = 0\), which gives \(x = 1\).
- Substitute the value \(x=1\) into the expression \(x^3 - \frac{1}{x^3}\).
- Calculate the result: \(1^3 - \frac{1}{1^3} = 1 - 1 = 0\).
| Equation | \(x^2 - 2x + 1 = 0\) |
|---|---|
| Factored Form | \((x-1)^2 = 0\) |
| Value of \(x\) | \(x = 1\) |
| Expression | \(x^3 - \frac{1}{x^3}\) |
| Substitution | \(1^3 - \frac{1}{1^3}\) |
| Final Value | \(1 - 1 = 0\) |
Revision Table: Key Concepts
| Concept | Description | Example |
|---|---|---|
| Quadratic Equation | An equation of the form \(ax^2 + bx + c = 0\), where \(a \neq 0\). | \(x^2 - 2x + 1 = 0\) |
| Perfect Square Trinomial | A trinomial that results from squaring a binomial, e.g., \((a-b)^2 = a^2 - 2ab + b^2\). | \(x^2 - 2x + 1 = (x-1)^2\) |
| Solving for \(x\) | Finding the value(s) of the variable that satisfy the equation. | For \((x-1)^2 = 0\), \(x=1\). |
| Expression Evaluation | Substituting a known value of a variable into an expression and calculating the result. | Evaluate \(x^3 - \frac{1}{x^3}\) when \(x=1\). |
Additional Information: Algebraic Identities
While not strictly needed for this particular problem since \(x=1\) is a simple value, sometimes solving expressions like \(x^3 - \frac{1}{x^3}\) where \(x + \frac{1}{x}\) or \(x - \frac{1}{x}\) is known involves algebraic identities.
One useful identity for expressions involving cubes is:
\(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)
If we let \(a=x\) and \(b=\frac{1}{x}\), then:
\(x^3 - \frac{1}{x^3} = (x - \frac{1}{x})(x^2 + x(\frac{1}{x}) + (\frac{1}{x})^2)\)
\(x^3 - \frac{1}{x^3} = (x - \frac{1}{x})(x^2 + 1 + \frac{1}{x^2})\)
Also, we know that \((x - \frac{1}{x})^2 = x^2 - 2x(\frac{1}{x}) + (\frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}\). So, \(x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2\).
Substituting this back into the identity for \(x^3 - \frac{1}{x^3}\):
\(x^3 - \frac{1}{x^3} = (x - \frac{1}{x})((x - \frac{1}{x})^2 + 2 + 1)\)
\(x^3 - \frac{1}{x^3} = (x - \frac{1}{x})((x - \frac{1}{x})^2 + 3)\)
This identity could be used if the value of \(x - \frac{1}{x}\) were easily obtainable from the initial equation. However, in this specific problem, finding \(x\) directly was the simplest approach.
Important Questions from Identities
(x - y) 3+ (y - z) 3+ (z - x) 3= ?
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- \((\sqrt{7} + \sqrt{9})(\sqrt{7} - \sqrt{9})\) is equal to:
If x + y = 5 and xy = 6, then find x 3+ y 3
If \(x = \sqrt3 + \sqrt2,\) then the value of \(x^2 + \frac{1}{x^2}\) is:
