If ω is cube root of 1 and α,β,γ are cube roots of p , then for any real numbers x,y,z what is the value of (xα+yβ+zγ)/(xβ+yγ+zα)?

Question

This is a problem I found while doing self study for the IITJEE entrance exam. It is a question from IITJEE 1990. How do I begin on this problem? The book says α = (p)^1/3, β = w(p)^1/3 and γ = (w^2)((p)^1/3). How did they get this? Shouldn't all three be equal to p^(1/3)? And I realize that 1, w and w^2 are cube roots of 1. However w and w^2 are not equal to one, or to each other. How can they be multiplied?

Answer 1

To see that $w\alpha$ and $w^2\alpha$ are also cube roots of $p$, just cube them and simplify using the laws of exponents: $$(w\alpha)^3 = w^3 \alpha^3 = 1 \cdot p = p $$ and $$(w^2\alpha)^3 = w^6 \alpha^3 = (w^3)^2 p = 1 \cdot p = p $$

Answer 2

Given that \begin{align} z_1^3 &=1 \\ z_2^3 &= p \end{align} then it is well known that solutions for $z_1$ are \begin{align} z_1:\left\{ \begin{array}{ll} 1\\ \omega=\frac{-1+i\sqrt{3}}{2}\\ \omega^2=\frac{-1-i\sqrt{3}}{2} \end{array} \right. \end{align} Now, write \begin{align} z_2^3 &= p \\ \implies z_2^3 &=1.z_1^3 \\ \implies \alpha &= \sqrt[3]{1}\sqrt[3]{p} \\ &= p^{\frac{1}{3}}\\ \beta &= \sqrt[3]{1}\sqrt[3]{p} \\ &= \omega p^{\frac{1}{3}} \\ \gamma &= \sqrt[3]{1}\sqrt[3]{p} \\ &= \omega^2 p^{\frac{1}{3}} \end{align}

Now let $$z=\frac{x \alpha + y \beta + z \gamma}{x \beta + y \gamma + z \alpha}$$

hence \begin{align} z &= \frac{x p^{\frac{1}{3}} + y \omega p^{\frac{1}{3}} + z \omega^2 p^{\frac{1}{3}}}{x \omega p^{\frac{1}{3}} + y \omega^2 p^{\frac{1}{3}} + z p^{\frac{1}{3}}}\\ &= \frac{x + y \omega + z \omega^2 }{x \omega + y \omega^2 + z }\\ &= \frac{\omega^2(x \omega + y \omega^2 + z) }{x \omega + y \omega^2 + z }\\ &= \omega^2 \end{align} Similalry by taking $\beta = \omega p^{\frac{1}{3}}$ and $\gamma = \omega^2 p^{\frac{1}{3}}$ you obtain $z=\omega$.