Question

If w be complex cube root of unity satisfying the relation $\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega}$= 2w² and $\frac{1}{a + \omega^{2}} + \frac{1}{b + \omega^{2}} + \frac{1}{c + \omega^{2}}$ = 2w then value of $\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}$is equal to-

• A. 2
• B. –2
• C. –1 + w²
• D. –1 + w

Answer 1

Answer: (A)

2

Answer 2

Sol. Given two relation show that the w and w² are the roots of $\frac{1}{a + x} + \frac{1}{b + x} + \frac{1}{c + x}$= $\frac{2}{x}$

which is cubic in x and we have to prove that 1 is roots of it

x S (b + x) (c + x) = 2(a + x) (b + x) (c + x)

Ž x[S bc + 2(Sa) x + 3x²] = 2[x³ + x² S a + x S ab + abc]

Ž 3x³ + 2x² S a + x S bc = 2x³ + 2x² S a + 2x S ab + 2abc

Ž x³ + 0. x² – x S ab – 2abc = 0

It is cubic in x whose roots are w, w², and

Let third root is a

Now w + w² + a = 0

Ž a = 1 (Q w + w² = –1)

Ž $\sum_{}^{}\frac{1}{a + 1}$ = $\frac{2}{1}$ = 2.`