Question

please answer this question​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given:

$\tt \implies \dfrac{x}{y} + \dfrac{y}{x} = - 1$

We have to find out the value of x³ - y³.

LCM of x and y is xy.

$\tt \implies \dfrac{ {x}^{2} + {y}^{2}}{xy} = - 1$

Move xy to right side. We get,

$\tt \implies{x}^{2} + {y}^{2} = -xy$

$\tt \implies{x}^{2} + xy + {y}^{2} = 0$

Now,

→ x³ - y³ = (x - y)(x² + xy + y²)

As x² + xy + y² = 0,

→ x³ - y³ = 0

$\texttt{\textsf{\large{\underline{Answer}:}}}$

• x³ - y³ = 0

$\texttt{\textsf{\large{\underline{Additionals}:}}}$

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a + b)³ = a³ + 3ab(a + b) + b³
• (a - b)³ = a³ - 3ab(a - b) - b³
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)