Solveeit Logo
Login

Question

Mathematics Question on argand plane

If α,β\alpha, \beta and γ\gamma are the roots of the equation x3−3x2+3x+7=0x^3 - 3x^2 + 3x + 7 = 0, and w is cube root of unity, then the value of α−1β−1+β−1γ−1+γ−1α−1\frac{\alpha-1}{\beta-1} + \frac{\beta-1}{\gamma-1} + \frac{\gamma-1}{\alpha-1} is equal to

A

3w23 w^2

B

3/w3/w

C

2w22w^2

D

none of these

Answer

3w23 w^2

Explanation

Solution

We have,
x3−3x2+3x+7=0x^{3}-3x^{2}+3x+7=0
(x+1)(x2−4x+7)=0\left(x + 1\right)\left(x^{2} - 4x + 7\right) = 0
x+1=0x+1=0 or x2−4x+7=0x^{2}-4x+7=0
x=−1x=-1 or x=4±16−282x=\frac{4 \pm\sqrt{16-28}}{2}
⇒x=−1\Rightarrow x=-1 or x=4±23i2x=\frac{4\pm2\sqrt{3}i}{2}
⇒x=−1\Rightarrow x=-1 or x=2±3ix=2 \pm\sqrt{3}i
⇒x=−1\Rightarrow x=-1 or x=1−2wx=1-2w, 1−2w,1−2w21-2w, 1-2w^{2}
[2w=−1+3i,2w2=−1−3i]\left[2w=-1+\sqrt{3}i, 2w^{2}=-1-\sqrt{3}i\right]
∴α=−1,β=1−2w\therefore \alpha=-1, \beta=1-2w and γ=1−2w2\gamma=1-2w^{2}
Now, α−1β−1+β−1γ−1+γ−1α−1\frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}
=−1−11−2w−1+1−2w−11−2w2−1=1−2w2−1−1−1=\frac{-1-1}{1-2w-1}+\frac{1-2w-1}{1-2w^{2}-1}=\frac{1-2w^{2}-1}{-1-1}
=22w+2w2w2+2w22=\frac{2}{2w}+\frac{2w}{2w^{2}}+\frac{2w^{2}}{2}
=1w+1w+w2=2w+w2=\frac{1}{w}+\frac{1}{w}+w^{2}=\frac{2}{w}+w^{2}
=2w2+w2=3w2=2w^{2}+w^{2}=3w^{2}