$$(x + y - 7)^2 + (y + 3x - 13)^2 = 0$$
The value of $(x^3 + y^3)$ is ________ (in integer).
Solving the Integer Equation
The given equation is: $ (x + y - 7)^2 + (y + 3x - 13)^2 = 0 $ Since the squares of real numbers are always non-negative, the sum of two squares can only be zero if each term is individually zero. Therefore, we have two separate equations:
- Equation 1: \( x + y - 7 = 0 \)
- Equation 2: \( y + 3x - 13 = 0 \)
Finding Integer Values for x and y
We need to solve this system of linear equations for the integers \(x\) and \(y\).
- From Equation 1, express \(y\) in terms of \(x\): $ y = 7 - x $
- Substitute this expression for \(y\) into Equation 2: $ (7 - x) + 3x - 13 = 0 $
- Simplify and solve for \(x\): $ 2x - 6 = 0 $ $ 2x = 6 $ $ x = 3 $
- Substitute the value of \(x = 3\) back into the expression for \(y\): $ y = 7 - 3 $ $ y = 4 $
The integer solutions are \(x = 3\) and \(y = 4\).
Calculating x³ + y³
Now, we calculate the value of \(x^3 + y^3\) using the found integer values:
$ x^3 + y^3 = 3^3 + 4^3 $ $ 3^3 = 3 \times 3 \times 3 = 27 $ $ 4^3 = 4 \times 4 \times 4 = 64 $ $ x^3 + y^3 = 27 + 64 $ $ x^3 + y^3 = 91 $The calculated value of \( (x^3 + y^3) \) is 91.
Important Questions from Algebra
- For positive non-zero real variables $x$ and $y$, if
$ln\left(\frac{x+y}{2}\right) = \frac{1}{2} [ln\left(x\right) + ln\left(y\right)]$
then, the value of $\frac{x}{y} + \frac{y}{x}$ is Given $f(x, y) = x^2 - 2xy + y^2$
The complete contour of the equation $f(x, y) = 1$ is described by the option(s) ___.
- If $pqr \neq 0$ and $p^{-x} = \frac{1}{q}$, $q^{-y} = \frac{1}{r}$, $r^{-z} = \frac{1}{p}$, what is the value of the product $xyz$?
- Two points $(4, p)$ and $(0, q)$ lie on a straight line having a slope of $3/4$. The value of $(p – q)$ is
- If $Pe^x = Qe^{-x}$ for all real values of $x$, which one of the following statements is true?
