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Q.

If $y + \frac{1}{y} = 1$ then find the value of $y^{3}$ .

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a

2

b

1

c

-1

d

0

answer is C.

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Detailed Solution

Given, equation is,

y + \frac{1}{y} = 1

……..(1)

y^{2} + 1 = y

y^{2} - y + 1 = 0

y^{2} = y - 1

……..(2)
It is a quadratic equation.
On compare it with the standard quadratic equation

a x^{2} + bx + c = 0,

we get,

a = 1, b = - 1 and c = 1

Put the values in the following quadratic formula,

x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

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

y = \frac{1 \pm \sqrt{3} i}{2}

Cube both sides of the equation,

y^{3} = {(\frac{1 \pm \sqrt{3} i}{2})}^{3}

y^{3} = \frac{(1 \pm \sqrt{3} i)^{3}}{8}

Also use the formula

(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)

(a - b)^{3} = a^{3} - b^{3} - 3 ab (a - b)

So, one value of

y^{3}

= \frac{1 + (\sqrt{3} i)^{3} + 3 \sqrt{3} i (1 + \sqrt{3} i)}{8}

= \frac{1 - (3 \sqrt{3} i) + 3 \sqrt{3} i - 9}{8}

[i^{2} = - 1]

= \frac{- 8}{8}

= - 1

Another value of

y^{3}

,

= \frac{1 - (\sqrt{3} i)^{3} - 3 \sqrt{3} i (1 - \sqrt{3} i)}{8}

= \frac{1 - (- 3 \sqrt{3} i) - 3 \sqrt{3} i - 9}{8}

= \frac{- 8}{8}

= - 1

Correct option is 3.

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