Algebraic manipulations

Change $\dfrac{1}{x}+\dfrac{1}{y}=1$ to the form $\dfrac{x+y}{xy}=1$ , or $x+y=xy$ . Knowing $x+y$ , we can determine that $xy=5$ . Now, we factor $x^3+y^3$ into $(x+y)(x^2+y^2-xy)$ . Expand $(x+y)^2=5^2=25$ to $x^2+y^2+2xy=25$ and substitute $xy$ to find $x^2+y^2+10=25$ , so $x^2+y^2=15$ . Now, we substitute to find that our answer is $(5)(15-5) = 5 \cdot 10 = \fbox{50}$

I like all other solutions from @Josh Speckman , @Alan Enrique Ontiveros Salazar , and @Mahdi Al-kawaz . I have an interesting approach that differs from your straightforward approach. We can start off the same, by condensing the second equation to $\frac{x+y}{xy}=1$ . Then, by substituting the first equation into the numerator, it appears that $\frac{5}{xy}=1 \Longrightarrow xy=5$ . Instead of now manipulating $x^3+y^3$ , remember Vieta's Formulas. Using these rules, we can construct a quadratic such that its roots are $x$ and $y$ . This is simply $a^2-5a+5=0$ . Now, we can use Newton's Sums to find $x^3+y^3$ . This is one of my favorite tricks, and applying this produces $\boxed{50}$ . Enjoy! :D

Rewrite the second equation by multiplying both sides with $xy$ : $xy=x+y$ Replace with the value of $x+y=5$ : $xy=5$ Now, rewrite $x^{3}+y^{3}$ as $(x+y)^{3}-3xy(x+y)$ , so: $x^{3}+y^{3}=(x+y)^{3}-3xy(x+y)$ And replace with the known values: $x^{3}+y^{3}=(5)^{3}-3(5)(5)$ $x^{3}+y^{3}=125-75$ $x^{3}+y^{3}=\boxed{50}$

$\frac {1}{x}+\frac{1}{y}=1 \Rightarrow xy=5$

$(x+y)^{3}=125$ $x^{3}+y^{3}=125-3xy(x+y)$ $x^{3}+y^{3}=125-75=\boxed {50}$

$\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{x+y}{xy}$ = 1, so x+y= xy =5 (1). Using binomial expansion (i.e. the 3rd row of pascals triangle), we produce the following identity: $x^{3}$ + $y^{3}$ = $(x+y)^{3}$ - 3 $x^{2}$ y - 3 $y^{2}$ ${x}$ = $(x+y)^{3}$ -3xy(x+y) = $5^{3}$ - (3X5X5) = 125-75 = $\boxed{50}$

Given:

$x+y=5$

$\frac{1}{x}+\frac{1}{y}=1$

Solving this equation, we get:

$\frac{x+y}{xy}=1$

Putting $x+y=5$ , we get:

$\frac{5}{xy}=1$

$xy=5$

Now,

$x^{3}+y^{3}=(x+y)^{3}-3xy(x+y)$

Substituting values, we get:

$x^{3}+y^{3}=5^{3}-3\times5\times5$

$x^{3}+y^{3}=125-75$

$x^{3}+y^{3}=50$

Thus, the answer is: $\boxed{x^{3}+y^{3}=50}$

X+Y=5, 1/X+1/Y=1 OR, (X+Y)/XY =1 OR, XY=5, (X+Y)^3 -3XY(X+Y) =X^3+Y^3 = 5^3-(3 5 5)=125-75=50

Here, x+y=5, and xy=5, (x+y)^3-3xy(x+y), =, 125-3 5 5 = 50

x+y=5 xy=x+y=5 (x+y)^3=125 x^3+3x^2y+3xy^2+y^3=125 x^3+y^3+3xy(x+y)=125 x^3+y^3+3 5 5=125 x^3+y^3=50

paragraph 1 x + y = 5 \frac{1}{x} + \frac{1}{y} = 1 \frac{x+y}{xy} = 1 xy = 5

paragraph 2 x^{2} + y^{2} = (x + y)^{2} - 2xy x^{2} + y^{2} = 5^{2} - 2(5) x^{2} + y^{2} = 25 - 10 x^{2} + y^{2} = 15

paragraph 3 x^{3} + y^{3} = (x + y)(x^{2} - xy + y^{2}) x^{2} + y^{2} = (5)(15-5) x^{2} + y^{2} = 5(10) x^{2} + y^{2} = 50

solved it's very simple .-.