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  • If \frac{x}{y}+\frac{y}{x}=-1 \left ( x,y\neq 0 \right ) the vlaue of x^{3}-y^{3} is (A) 1 (B)-1 (C)0 (D) \frac{1}{2}

If   \frac{x}{y}+\frac{y}{x}=-1  \left ( x,y\neq 0 \right ) the value of x^{3}-y^{3} is

(A) 1     (B) -1     (C)0    (D) \frac{1}{2}

Answers (1)

(C) 0

Solution

Given \frac{x}{y}+\frac{y}{x}=-1  \left ( x,y\neq 0 \right )

Simplifying the above equation, we have

\\\frac{x^{2}+y^{2}}{xy}=-1\\ x^{2}+y^{2}=-xy\\ x^{2}+y^{2}+xy=0 \cdots \cdots \cdots (1)\\ x^{3}-y^{3}=(x-y)(x^{2}+y^{2}+xy)\cdots \cdots \cdots (2)\\ \left \{ \because a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab) \right \}

  Put the value of equation 1 in 2 and we get

\\x^{3}-y^{3}=(x-y)(0)\\ x^{3}-y^{3}=0

Hence, option C is correct.

 

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infoexpert24

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