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Find the value of x and y using the equations $x^{5} - y^{5} = 992$ and $x - y = 2$ .

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x = 4, - 2, 1 \pm \sqrt{- 11} and y = 2, - 4, - 1 \pm \sqrt{- 11}

respectively

x = 3, - 5, 2 \pm \sqrt{- 11} and y = 5, - 6, - 2 \pm \sqrt{- 11}

respectively

x = 10, - 2, 1 - \sqrt{- 11} and y = 6, 4, 1 \pm \sqrt{- 12}

respectively

x = 6, - 8, 3 - \sqrt{- 11} and y = 7, - 2, - 5 \pm \sqrt{- 21}

respectively

answer is B.

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Detailed Solution

Given equation,

x^{5} - y^{5} = 992 \dots \dots \dots (i)

x - y = 2 \dots \dots \dots (ii)

Solving equation (ii), we get

x - y = 2

\Rightarrow x = y + 2 \dots \dots \dots (iii)

Substituting value of (iii) in equation (i), we get

(y + 2)^{5} - y^{5} = 992 \dots \dots \dots (iv)

\Rightarrow (y + 2)^{3} (y + 2)^{2} - y^{5} = 992

\Rightarrow (y^{3} + 8 + 6 y^{2} + 12 y) (y^{2} + 4 + 4 y) - y^{5} = 992

\Rightarrow y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32 - y^{5} = 992

\Rightarrow 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y = 960

\Rightarrow y^{4} + 4 y^{3} + 8 y^{2} + 8 y - 96 = 0

From hit and trial method.
Put y = 2

16 + 32 + 32 + 16 - 96 = 0

Hence, y − 2 is one of the root of equation.

\Rightarrow (y - 2) (y^{3} + 6 y^{2} + 20 y + 48) = 0

Thus y−2 = 0
Also,

y^{3} + 6 y^{2} + 20 y + 48 = 0

Put y = −4

- 64 + 96 - 80 + 48 = 0

Hence, y + 4 also a root of equation.

\Rightarrow (y + 4) (y^{2} + 2 y + 12) = 0

Thus y + 4 = 0
Now in quadratic equation,

y^{2} + 2 y + 12 = 0

from quadratic formula, we get,

y = \frac{- 2 \pm \sqrt{4 - 4 (1) (12)}}{2}

= \frac{- 2 \pm \sqrt{- 44}}{2}

y = - 1 \pm \sqrt{- 11}

……. (iv)
Thus, value of y are,

y = 2, - 4, - 1 \pm \sqrt{- 11}

Now, we have,

x = y + 2

So, the values of x at

y = 2, - 4, - 1 \pm \sqrt{- 11}

\Rightarrow x = 4, - 2, 1 \pm \sqrt{- 11}

respectively.
Thus,
Option 2 is correct.

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Find the value of x and y using the equations $x^{5} - y^{5} = 992$ and $x - y = 2$ .