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# If $x+y=5$ and $xy=4$, find ${{x}^{3}}+{{y}^{3}}$.

Last updated date: 13th Jun 2024
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Hint: In solving the solution of this question, we will first substitute from both the equations. We will write the equation $x+y=5$ by putting the term y in the left side of equation and the in the right side of the equation, we will put the term y and the constant term 5. From here, we will take the value of y and put this value of y in the equation $xy=4$ and find the value of x. And, from there we will find the value of y by putting the value of x in any of the equations. After that, we will put the value of x and y in the term ${{x}^{3}}+{{y}^{3}}$, then we will get the answer.

Complete step by step solution:
Let us solve this question.
In this question, we have given two equations that are $x+y=5$ and $xy=4$.
Here, we will first solve the equation $x+y=5$.
We will take the value of x to the right side of equation $x+y=5$
So, the equation $x+y=5$ can also be written as
$\Rightarrow y=5-x$
Now, we will put this value of y (that is $y=5-x$) in the equation $xy=4$.
After putting the value of y in the equation $xy=4$, then we will get
$\Rightarrow x\left( 5-x \right)=4$
We can write the above equation as
$\Rightarrow 5x-{{x}^{2}}=4$
We can write the above equation as
$\Rightarrow {{x}^{2}}-5x+4=0$
As we know that
If the equation is $a{{x}^{2}}+bx+c=0$, then the value of x will be
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Hence, the value of x in the equation ${{x}^{2}}-5x+4=0$ will be
$x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 1\times 4}}{2\times 1}$
The above can also be written as
$x=\dfrac{5\pm \sqrt{25-16}}{2}=\dfrac{5\pm \sqrt{9}}{2}=\dfrac{5\pm 3}{2}$
Hence, the value of x will be $\dfrac{5+3}{2}$ and $\dfrac{5-3}{2}$
Or, we can say that the values of x are 4 and 1.
If we will put the value of x as 4 in the equation $xy=4$, then the value of y will be 1.
If we will put the value of x as 1 in the equation $xy=4$, then the value of y will be 4.
For the value of x and y as 4 and 1 respectively, then the value of the term ${{x}^{3}}+{{y}^{3}}$ will be ${{4}^{3}}+{{1}^{3}}=65$
For the value of x and y as 1 and 4 respectively, then the value of the term ${{x}^{3}}+{{y}^{3}}$ will be ${{1}^{3}}+{{4}^{3}}=65$
Hence, we get that the value of the term ${{x}^{3}}+{{y}^{3}}$ is 65.

Note:
We should know how to substitute a variable from the two equations, so that we can solve this type of question easily. We should know that if we have a given a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$, then the value of x for this type of equation will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
If we want to cross check if our answer is correct, then we can check the equations $x+y=5$ and $xy=4$ are satisfying or not by putting the value of x and y,
If we put the value of x as 4 or 1 and y as 1 or 4 in the equation $x+y=5$ , then the equation will be like: 5=5, hence it satisfies.
And, if we put the value of x as 4 or 1 and y as 1 or 4 in the equation $xy=4$, then the equation will be like: 4=4, hence it satisfies both the equation and says that answers are correct.