Question

Pls find the answer for question 2

Answer 1

Answer:

c

Step-by-step explanation:

x³ - y³ ← is a difference of cubes and factors as

x³ - y³ = (x - y)(x² + xy + y²)

Given

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

Multiply through by xy to clear the fractions

x² + y² = - xy ← substitute into second factor of expansion

x³ - y³ = (x - y)(- xy + xy) = (x - y) × 0 = 0 → c

Answer 2

Answer:

The answer is C. 0

Step-by-step explanation:

since

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

we multiply both sides by xy to cancel their denominators and multiply -1 by xy

so we have

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

we get our answer as

${x}^{2} + {y}^{2} = - xy$

we were given the difference of cubes that is

${x}^{3} - {y}^{3}$

which is =

$(x - y)( {x}^{2} + xy + {y}^{2} )$

so since,

${x}^{2} + {y}^{2} = - xy$

we substitute,

$(x - y)( - xy + xy) = (x - y)(0) = 0$