Question

so what would this be

x^-1 + y^-1
______________
x^-3 +y^-3

Answer 1

\[\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^3}+\frac{1}{y^3}}\]

remember
\[a^3+b^3=(a+b)(a^2-ab+b^2)\]

Answer 2

\[\frac{1}{x^3}+\frac{1}{y^3}=(\frac{1}{x}+\frac{1}{y})((\frac{1}{x})^2-\frac{1}{x} \cdot \frac{1}{y}+(\frac{1}{y})^2)\]

Answer 3

do you see that the numerator and denominator have common factors?

Answer 4

so now we have
\[\frac{1}{\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}}\]

Answer 5

so multiply by the reciprocal?

Answer 6

multiply top and bottom by
\[x^2y^2\]
to clear the fractions on bottom

Answer 7

why x^2 y^2

Answer 8

\[\frac{x^2y^2}{(x^2y^2)(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2})}=\frac{x^2y^2}{y^2-xy+x^2}\]

Answer 9

because we have x^2 and y^2 in the bottoms of the denominator

Answer 10

lcd?

Answer 11

the lcd of x^2,xy,y^2 is x^2y^2

Answer 12

ok

Answer 13

so is that the answer

Answer 14

not the lcd i mean the fraction

Answer 15

can it be simplified