If x + \(\frac{1}{x}\) = 7, then the value of x6 + \(\frac{1}{x^6}\) is:
103682
Finding the Value of x^6 + 1/x^6 Given x + 1/x
This question asks us to find the value of an algebraic expression, \(x^6 + \frac{1}{x^6}\), given the value of a simpler expression, \(x + \frac{1}{x}\). We are given that \(x + \frac{1}{x} = 7\). To find \(x^6 + \frac{1}{x^6}\), we can use algebraic identities by either squaring or cubing the initial expression step by step.
Step-by-Step Calculation Methods
We can solve this problem using a couple of common algebraic approaches. Both methods involve using the given equation \(x + \frac{1}{x} = 7\) and applying squaring or cubing identities repeatedly.
Method 1: Squaring then Cubing
This method first finds \(x^2 + \frac{1}{x^2}\) by squaring the given equation, and then finds \(x^6 + \frac{1}{x^6}\) by cubing the expression for \(x^2 + \frac{1}{x^2}\).
Step 1: Find the value of \(x^2 + \frac{1}{x^2}\).
Start with the given equation:
\[x + \frac{1}{x} = 7\]Square both sides of the equation:
\[\left(x + \frac{1}{x}\right)^2 = 7^2\]Using the identity \((a+b)^2 = a^2 + 2ab + b^2\), where \(a=x\) and \(b=\frac{1}{x}\):
\[x^2 + 2\left(x\right)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 49\] \[x^2 + 2 + \frac{1}{x^2} = 49\]Subtract 2 from both sides:
\[x^2 + \frac{1}{x^2} = 49 - 2\] \[x^2 + \frac{1}{x^2} = 47\]Step 2: Find the value of \(x^6 + \frac{1}{x^6}\) by cubing \(x^2 + \frac{1}{x^2}\).
We now have \(x^2 + \frac{1}{x^2} = 47\). Let \(y = x^2\) and \(\frac{1}{y} = \frac{1}{x^2}\). We want to find \(y^3 + \frac{1}{y^3}\).
Cube both sides of the equation \(x^2 + \frac{1}{x^2} = 47\):
\[\left(x^2 + \frac{1}{x^2}\right)^3 = 47^3\]Using the identity \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\), where \(a=x^2\) and \(b=\frac{1}{x^2}\):
\[\left(x^2\right)^3 + \left(\frac{1}{x^2}\right)^3 + 3\left(x^2\right)\left(\frac{1}{x^2}\right)\left(x^2 + \frac{1}{x^2}\right) = 47^3\] \[x^6 + \frac{1}{x^6} + 3(1)\left(x^2 + \frac{1}{x^2}\right) = 47^3\]Substitute the value \(x^2 + \frac{1}{x^2} = 47\) into the equation:
\[x^6 + \frac{1}{x^6} + 3(47) = 47^3\] \[x^6 + \frac{1}{x^6} + 141 = 47^3\]Calculate \(47^3\):
\[47^2 = 2209\] \[47^3 = 47 \times 2209 = 103823\]Substitute the value of \(47^3\) back into the equation:
\[x^6 + \frac{1}{x^6} + 141 = 103823\]Subtract 141 from both sides:
\[x^6 + \frac{1}{x^6} = 103823 - 141\] \[x^6 + \frac{1}{x^6} = 103682\]Method 2: Cubing then Squaring
This method first finds \(x^3 + \frac{1}{x^3}\) by cubing the given equation, and then finds \(x^6 + \frac{1}{x^6}\) by squaring the expression for \(x^3 + \frac{1}{x^3}\).
Step 1: Find the value of \(x^3 + \frac{1}{x^3}\).
Start with the given equation:
\[x + \frac{1}{x} = 7\]Cube both sides of the equation:
\[\left(x + \frac{1}{x}\right)^3 = 7^3\]Using the identity \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\), where \(a=x\) and \(b=\frac{1}{x}\):
\[x^3 + \left(\frac{1}{x}\right)^3 + 3\left(x\right)\left(\frac{1}{x}\right)\left(x + \frac{1}{x}\right) = 7^3\] \[x^3 + \frac{1}{x^3} + 3(1)\left(x + \frac{1}{x}\right) = 7^3\]Substitute the value \(x + \frac{1}{x} = 7\) into the equation:
\[x^3 + \frac{1}{x^3} + 3(7) = 7^3\] \[x^3 + \frac{1}{x^3} + 21 = 343\]Subtract 21 from both sides:
\[x^3 + \frac{1}{x^3} = 343 - 21\] \[x^3 + \frac{1}{x^3} = 322\]Step 2: Find the value of \(x^6 + \frac{1}{x^6}\) by squaring \(x^3 + \frac{1}{x^3}\).
We now have \(x^3 + \frac{1}{x^3} = 322\). Let \(z = x^3\) and \(\frac{1}{z} = \frac{1}{x^3}\). We want to find \(z^2 + \frac{1}{z^2}\).
Square both sides of the equation \(x^3 + \frac{1}{x^3} = 322\):
\[\left(x^3 + \frac{1}{x^3}\right)^2 = 322^2\]Using the identity \((a+b)^2 = a^2 + 2ab + b^2\), where \(a=x^3\) and \(b=\frac{1}{x^3}\):
\[\left(x^3\right)^2 + 2\left(x^3\right)\left(\frac{1}{x^3}\right) + \left(\frac{1}{x^3}\right)^2 = 322^2\] \[x^6 + 2 + \frac{1}{x^6} = 322^2\]Calculate \(322^2\):
\[322^2 = 103684\]Substitute the value of \(322^2\) back into the equation:
\[x^6 + 2 + \frac{1}{x^6} = 103684\]Subtract 2 from both sides:
\[x^6 + \frac{1}{x^6} = 103684 - 2\] \[x^6 + \frac{1}{x^6} = 103682\]Both methods yield the same result. The value of \(x^6 + \frac{1}{x^6}\) is 103682.
Revision Table: Key Values
| Expression | Value | Identity Used |
|---|---|---|
| \(x + \frac{1}{x}\) | 7 | Given |
| \(x^2 + \frac{1}{x^2}\) | 47 | \(\left(x + \frac{1}{x}\right)^2 - 2\) |
| \(x^3 + \frac{1}{x^3}\) | 322 | \(\left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)\) |
| \(x^6 + \frac{1}{x^6}\) | 103682 | \(\left(x^3 + \frac{1}{x^3}\right)^2 - 2\) or \(\left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2 + \frac{1}{x^2}\right)\) |
Additional Information on Algebraic Identities
Problems involving expressions like \(x^n + \frac{1}{x^n}\) where \(n\) is a power often rely on a few fundamental algebraic identities. Understanding these identities helps in quickly solving such problems.
- For finding the square: If \(x + \frac{1}{x} = k\), then \(x^2 + \frac{1}{x^2} = k^2 - 2\). This is derived from \(\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2\).
- For finding the cube: If \(x + \frac{1}{x} = k\), then \(x^3 + \frac{1}{x^3} = k^3 - 3k\). This is derived from \(\left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3x\frac{1}{x}\left(x + \frac{1}{x}\right) = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right)\).
- To find higher powers like \(x^6 + \frac{1}{x^6}\), we can apply these identities sequentially. \(x^6 + \frac{1}{x^6} = \left(x^3\right)^2 + \left(\frac{1}{x^3}\right)^2\), so if we know \(x^3 + \frac{1}{x^3}\), we can find the square of this term and subtract 2. Alternatively, \(x^6 + \frac{1}{x^6} = \left(x^2\right)^3 + \left(\frac{1}{x^2}\right)^3\), so if we know \(x^2 + \frac{1}{x^2}\), we can find the cube of this term and subtract 3 times the term itself.
These identities make it easier to compute higher powers without explicitly finding the value of \(x\).
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