Question

If x/y+y/x=1 , then find the value of x^3-y^3.

Answer 1

my friend x/y+y/x=-1

first let we take LCM of given equation

${x}^{2} + {y }^{2} \div xy = 1$

therefore,
${x +}^{2} + {y}^{2} = xy$
we know that,
${x}^{3} - {y}^{3} = (x - y)( {x }^{2} + {y}^{2} + xy)$
substituting the values,
${x}^{3} - {y}^{3} = (x - y)(0)$
therefore the answer is 0

Answer 2

x^3-y^3=(x-y) (x^2+y^2-xy)...1
x/y +y/x =1
x^2+y^2/xy =1
x^2+y^2=xy
x^2+y^2-xy=0
multiply both side by (x-y)
(x-y) (x^2+y^2-xy)=0(x-y)
so,(x-y) (x^2+y^2-xy)=0
so,x^3-y^3=0
hence proved.