Question

If $\dfrac{x}{y} + \dfrac{y}{x} = - 1$, $x \ne 0$, $y \ne 0$, then the value of ${x^3} - {y^3}$.

Answer 1

Hint: We have to find the value of ${x^3} - {y^3}$. Write ${x^3} - {y^3}$ in the factored form as ${x^3} - {y^3} = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)$. Then, simplify the given expression by adding and cross-multiplying to get the value of \[{x^2} + {y^2} + xy\]. Then, substitute its value in the expression to find the value of ${x^3} - {y^3}$.


Complete step-by-step answer:
We have to find the value of ${x^3} - {y^3}$
We know that ${x^3} - {y^3} = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)$ eqn. (1)
But, we are given that $\dfrac{x}{y} + \dfrac{y}{x} = - 1$
Which implies,
\[\dfrac{{{x^2} + {y^2}}}{{xy}} = - 1\], $x \ne 0$, $y \ne 0$
On cross multiplying, we will get,
$
  {x^2} + {y^2} = - xy \\
   \Rightarrow {x^2} + {y^2} + xy = 0 \\
$
We will substitute the above value of the expression in the equation (1)
This implies,
${x^3} - {y^3} = \left( {x - y} \right)\left( 0 \right)$
Which gives the value of ${x^3} - {y^3}$ as 0.

Note: Students must know the factored form of the expression, ${x^3} - {y^3}$. Also, students must know how to add unlike fractions. Note, that there cannot be any fraction with denominator 0, that’s why $x \ne 0$, $y \ne 0$ is an important condition for solving this question.