Question

Question If X/Y + Y/X = -1 (x,y # 0) then value of X^3-Y^3 is

Answer 1

ANSWER:

• Value of (x³-y³) = 0

GIVEN:

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

TO FIND:

Value of (x³-y³)

SOLUTION:

$\implies \: \frac{x}{y} + \frac{y}{x} = ( - 1) \\ \\ \implies \: \frac{x {}^{2} + y {}^{2} }{xy} = ( - 1) \\ \\ \implies \: x {}^{2} + y {}^{2} = - xy \\ \\ \implies \: x {}^{2} + y {}^{2} + xy = 0 \: \: \: ....(i)$

Formula

x³-y³ = (x-y)(x²+y²+xy)

Putting (x²+y²+xy)= 0. from eq (i)

=> x³-y³ = (x-y)*0

=> x³-y³ = 0

Value of (x³-y³) = 0

NOTE:

some important formulas:

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

Answer 2

Answer: 0

Step-by-step explanation:

TO FIND:

Value of (x³-y³)

SOLUTION:

Formula

x³-y³ = (x-y)(x²+y²+xy)

Putting (x²+y²+xy)= 0. from eq (i)

=> x³-y³ = (x-y)*0

=> x³-y³ = 0

Value of (x³-y³) = 0

hope it helps you✌