Math, asked by rishiraam, 5 months ago

If a, b, c, d are real numbers and
 \omega
is a non
real cube root of unity such that
 \frac{1}{a +  \omega}  +  \frac{1}{b +  \omega}  +  \frac{1}{c +  \omega}  +  \frac{1}{d +  \omega} =  \frac{2}{\omega}
then
 \frac{1}{abc}  +  \frac{1}{bcd}  +  \frac{1}{cda}  +  \frac{1}{dab}
equals :​

Answers

Answered by anukriti72
3

Answer:

Since,

w

2

=

1

w

and

w

=

1

w

2

Given relation may be rewritten as

1

a

+

w

+

1

b

+

w

+

1

c

+

w

=

2

w

and

1

a

+

w

2

+

1

b

+

w

2

+

1

c

+

w

2

=

2

w

2

Clearly

w

and

w

2

are the roots of

1

a

+

x

+

1

b

+

x

+

1

c

+

x

=

2

x

(

b

+

x

)

(

c

+

x

)

+

(

a

+

x

)

(

c

+

x

)

+

(

a

+

x

)

(

b

+

x

)

(

a

+

x

)

(

b

+

x

)

(

c

+

x

)

=

2

x

x

[

3

x

2

+

2

(

a

+

b

+

c

)

x

+

b

c

+

c

a

+

a

b

]

=

2

[

a

b

c

+

(

b

c

+

c

a

+

a

b

)

x

+

(

a

+

b

+

c

)

x

2

+

x

3

]

x

3

(

b

c

+

c

a

+

a

b

)

x

2

a

b

c

=

0

If

α

is the third root of this equation then sum of the roots

α

+

w

+

w

2

=

0

α

=

1

Hence,

1

is the root of equation, we get

1

a

+

1

+

1

b

+

1

+

1

c

+

1

=

2

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