Question

$\frac{x}{y} + \frac{y}{x} = - 1 \:$
If x,y are not equal to 0 then what is the value of
$x {}^{3} - y {}^{3}$


​

Answer 1

Solution

GIVEN :-

• x/y + y/x = - 1 --------------(1)

Find :-

• Value of ( x³ - y³)

Explanation

Using Formula

★ ( a³ - b³) = (a-b)(a²+b²+ab)

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Now by equ(1)

==> x/y + y/x = -1

==> (x²+y²)/xy = -1

==> (x²+y²) = - xy

==> (x²+y²+xy) = 0 ----------(2)

_______________________

Now, Calculate Value of ( x³ - y³)

==> (x³ - y³) = (x-y) (x²+y²+xy)

Keep Value by equ(2)

==> (x³ - y³) = (x-y) * 0

==> (x³ - y³) = 0

_______________________

Hence

• Value of (x³-y³) = 0

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Answer 2

★ QUESTION ★

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

If x,y are not equal to 0 then what is the value of

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

$\rule{305}2$

◇ Solution ◇

$\longrightarrow \frac{x}{y}+\frac{y}{x}=-1 \\ \longrightarrow \frac{x^2+y^2}{xy}= -1 \\ \longrightarrow x^{2}+y^{2}=-xy \\ \longrightarrow x^{2}+y^{2}+xy=0---eq(1)$

★By using identity★

$\longrightarrow (x^{3}-y^{3})=(x-y)(x^2+xy+y^2)$

★ According to question ★

→ we have to find the value of $\longrightarrow (x^{2}-y^{2})$

→[by equation (1)––––

$\longrightarrow (x^{2}-y^{2}=(x-y)(x^2+xy+y^2) \\ \longrightarrow x^{3}-y^{3}=(x-y)( 0 ) \\ \longrightarrow x^{3}-y^{3}=0$

Answer = 0