If x + y = 5 and xy = 6, then find x 3+ y 3
35
Finding x³ + y³ Given x + y and xy
This problem requires us to calculate the value of \(x^3 + y^3\) given the sum of \(x\) and \(y\), and the product of \(x\) and \(y\).
We are provided with the following information:
- The sum: \(x + y = 5\)
- The product: \(xy = 6\)
We need to find the value of \(x^3 + y^3\).
Applying the Correct Algebraic Identity
To solve this, we can use a common algebraic identity that relates the sum of cubes (\(x^3 + y^3\)) to the sum (\(x+y\)) and product (\(xy\)) of the variables. The relevant identity is:
\(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\)
This form is ideal because we are directly given the values for \((x+y)\) and \(xy\).
Step-by-Step Calculation of x³ + y³
Let's substitute \(a=x\) and \(b=y\) into the identity:
\(x^3 + y^3 = (x+y)^3 - 3xy(x+y)\)
Now, substitute the given values \(x+y = 5\) and \(xy = 6\) into this equation:
\(x^3 + y^3 = (5)^3 - 3(6)(5)\)
First, calculate the term \((5)^3\):
\(5^3 = 5 \times 5 \times 5 = 25 \times 5 = 125\)
Next, calculate the term \(3(6)(5)\):
\(3 \times 6 = 18\)
\(18 \times 5 = 90\)
Now, substitute these results back into the equation for \(x^3 + y^3\):
\(x^3 + y^3 = 125 - 90\)
Finally, perform the subtraction:
\(125 - 90 = 35\)
Result
The value of \(x^3 + y^3\) is 35.
Revision Table: Useful Algebraic Identities
| Identity Name | Formula |
|---|---|
| Sum of Cubes | \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\) |
| Sum of Cubes (Factored) | \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) |
| Difference of Cubes | \(a^3 - b^3 = (a-b)^3 + 3ab(a-b)\) |
| Difference of Cubes (Factored) | \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) |
| Square of a Sum | \((a+b)^2 = a^2 + 2ab + b^2\) |
| Square of a Difference | \((a-b)^2 = a^2 - 2ab + b^2\) |
Additional Information: Connecting Identities
The problem can also be approached using the factored form of the sum of cubes identity: \(x^3 + y^3 = (x+y)(x^2 - xy + y^2)\).
To use this identity, we first need to find the value of \(x^2 + y^2\).
We know that \((x+y)^2 = x^2 + 2xy + y^2\).
Rearranging this identity to solve for \(x^2 + y^2\), we get:
\(x^2 + y^2 = (x+y)^2 - 2xy\)
Substitute the given values \(x+y=5\) and \(xy=6\):
\(x^2 + y^2 = (5)^2 - 2(6)\)
\(x^2 + y^2 = 25 - 12\)
\(x^2 + y^2 = 13\)
Now, substitute the values of \(x+y\), \(xy\), and \(x^2 + y^2\) into the factored form of the sum of cubes identity:
\(x^3 + y^3 = (x+y)(x^2 - xy + y^2)\)
\(x^3 + y^3 = (5)(13 - 6)\)
\(x^3 + y^3 = (5)(7)\)
\(x^3 + y^3 = 35\)
This alternative method confirms our result and demonstrates how different algebraic identities are related and can be used to solve similar problems.
Important Questions from Identities
(x - y) 3+ (y - z) 3+ (z - x) 3= ?
If \(x + \left( {\frac{1}{x}} \right) = 12\) and \({x^2} - \frac{1}{{{x^2}}} = 50\) , then the value of \({x^4} - \frac{1}{{{x^4}}} \) is:
- \((\sqrt{7} + \sqrt{9})(\sqrt{7} - \sqrt{9})\) is equal to:
If x satisfies the equation x 2 - 2x + 1 = 0, then the value of \(\rm x^3 - \frac{1}{x^3}\) is:
If \(x = \sqrt3 + \sqrt2,\) then the value of \(x^2 + \frac{1}{x^2}\) is:
