If 2x 2- 8x - 1 = 0, then what is the value of \(\rm 8x^3 - \frac{1}{x^3}\) ?
560
Solving the Equation and Finding the Expression Value
We are given the equation \(2x^2 - 8x - 1 = 0\) and asked to find the value of the expression \(8x^3 - \frac{1}{x^3}\).
Step 1: Manipulate the Given Equation
Let's start by rearranging the given equation to find a relationship involving \(x\) and \(\frac{1}{x}\). The equation is:
\[2x^2 - 8x - 1 = 0\]Assuming \(x \neq 0\) (if \(x=0\), the original equation becomes \(2(0)^2 - 8(0) - 1 = -1 \neq 0\), so \(x\) cannot be zero), we can divide the entire equation by \(x\):
\[\frac{2x^2}{x} - \frac{8x}{x} - \frac{1}{x} = \frac{0}{x}\] \[2x - 8 - \frac{1}{x} = 0\]Now, isolate the terms with \(x\) and \(\frac{1}{x}\):
\[2x - \frac{1}{x} = 8\]This gives us a useful relationship: the value of \(2x - \frac{1}{x}\) is 8.
Step 2: Identify the Target Expression and Relevant Identity
The expression we need to evaluate is \(8x^3 - \frac{1}{x^3}\). We can rewrite this as:
\[8x^3 - \frac{1}{x^3} = (2x)^3 - \left(\frac{1}{x}\right)^3\]This expression is in the form of \(a^3 - b^3\), which is the difference of cubes. The algebraic identity for the difference of cubes is:
\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]In our case, \(a = 2x\) and \(b = \frac{1}{x}\). Substituting these into the identity:
\[(2x)^3 - \left(\frac{1}{x}\right)^3 = \left(2x - \frac{1}{x}\right)\left((2x)^2 + (2x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2\right)\] \[8x^3 - \frac{1}{x^3} = \left(2x - \frac{1}{x}\right)\left(4x^2 + 2 + \frac{1}{x^2}\right)\]So, to find the value of \(8x^3 - \frac{1}{x^3}\), we need the values of \(2x - \frac{1}{x}\) and \(4x^2 + \frac{1}{x^2}\).
Step 3: Calculate the Required Terms
From Step 1, we already found that \(2x - \frac{1}{x} = 8\).
Now let's find the value of \(4x^2 + \frac{1}{x^2}\). We can get this by squaring the expression \(2x - \frac{1}{x}\):
\[\left(2x - \frac{1}{x}\right)^2 = (2x)^2 - 2(2x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2\] \[\left(2x - \frac{1}{x}\right)^2 = 4x^2 - 4 + \frac{1}{x^2}\]We know \(2x - \frac{1}{x} = 8\), so substitute this value:
\[(8)^2 = 4x^2 - 4 + \frac{1}{x^2}\] \[64 = 4x^2 - 4 + \frac{1}{x^2}\]Add 4 to both sides to find the value of \(4x^2 + \frac{1}{x^2}\):
\[64 + 4 = 4x^2 + \frac{1}{x^2}\] \[68 = 4x^2 + \frac{1}{x^2}\]So, the value of \(4x^2 + \frac{1}{x^2}\) is 68.
Step 4: Substitute and Calculate the Final Value
Now we have the values for the terms needed in the identity:
- \(2x - \frac{1}{x} = 8\)
- \(4x^2 + \frac{1}{x^2} = 68\)
Recall the identity expression:
\[8x^3 - \frac{1}{x^3} = \left(2x - \frac{1}{x}\right)\left(4x^2 + 2 + \frac{1}{x^2}\right)\]Substitute the values we found:
\[8x^3 - \frac{1}{x^3} = (8)(68 + 2)\] \[8x^3 - \frac{1}{x^3} = (8)(70)\] \[8x^3 - \frac{1}{x^3} = 560\]The value of \(8x^3 - \frac{1}{x^3}\) is 560.
Summary of Calculation
| Step | Calculation/Relation | Result |
|---|---|---|
| Start Equation | \(2x^2 - 8x - 1 = 0\) | |
| Divide by x | \(2x - 8 - \frac{1}{x} = 0\) | |
| Rearrange | \(2x - \frac{1}{x} = 8\) | \(2x - \frac{1}{x} = 8\) |
| Square \(2x - \frac{1}{x}\) | \((2x - \frac{1}{x})^2 = 8^2\) | \(4x^2 - 4 + \frac{1}{x^2} = 64\) |
| Find \(4x^2 + \frac{1}{x^2}\) | \(4x^2 + \frac{1}{x^2} = 64 + 4\) | \(4x^2 + \frac{1}{x^2} = 68\) |
| Apply Identity | \(8x^3 - \frac{1}{x^3} = (2x - \frac{1}{x})(4x^2 + \frac{1}{x^2} + 2)\) | |
| Substitute Values | \((8)(68 + 2)\) | \((8)(70)\) |
| Final Value | \(8 \times 70\) | 560 |
The calculated value 560 matches option 2.
Revision Table: Key Algebraic Identities
| Identity | Formula |
|---|---|
| Square of a difference | \((a-b)^2 = a^2 - 2ab + b^2\) |
| Difference of cubes | \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) |
| Sum of cubes | \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) |
Additional Information: Solving Polynomial Expressions
Solving problems like finding the value of \(8x^3 - \frac{1}{x^3}\) from a quadratic equation \(2x^2 - 8x - 1 = 0\) often involves:
- Manipulating the given equation to find relationships between \(x\) and inverse terms (\(\frac{1}{x}\)).
- Recognizing the target expression as a standard algebraic form, like a difference or sum of cubes, or a squared term.
- Applying appropriate algebraic identities to simplify the expression.
- Calculating intermediate values based on the manipulated equation.
- Substituting these values back into the expanded form of the target expression.
For quadratic equations like \(ax^2 + bx + c = 0\), if \(x \neq 0\), dividing by \(x\) gives \(ax + b + \frac{c}{x} = 0\), which can lead to terms like \(x + \frac{c}{ax}\) or \(x - \frac{|c|}{|a|x}\) depending on signs. These forms are useful when the target expression involves cubes or squares of \(x\) and \(\frac{1}{x}\).
Important Questions from Identities
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If \(\rm x+ \frac{1}{x} = 4,\) then the value of \(\rm x^5 + \frac{1}{x^5}\) is:
If x + y = 1, then what is the value of x 3+ 3xy + y 3?
