But I'm Only Given One Equation!

Algebra Level 2

If $xy(x+y) = 1$ , find the value of $\dfrac1{x^3y^3} - x^3 -y^3$ .

1 solution

Goh Choon Aik
Apr 13, 2016

xy(x+y) = 1

If we manipulate this equation, we find that $\frac {1}{x^{3} y^{3}} = (x+y)^ {3}$

Expanding the right side gives $\frac {1}{x^{3} y^{3}} = x^ {3} + y^ {3} +3xy(x+y)$

Then we find

$\frac {1}{x^{3} y^{3}} - x^ {3} - y^ {3} = 3xy(x+y)$

And that $3xy(x+y) = 3 \times 1$

Therefore the solution is 3.

How do you manipulate?

Angel White - 5 years, 2 months ago

We first divide both sides by xy.

$xy(x+y) = 1,$

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

Next we cube both sides:

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

$(x+y)^ {3} = \frac {1}{(xy)^ {3}}$

Goh Choon Aik - 5 years, 2 months ago

okay i understand

Angel White - 5 years, 2 months ago