Free algebra! part 2

Algebra Level 3

$\frac { 7+4\omega }{ 7{ \omega }^{ 2 }+4 } +\frac { 5-3{ \omega }^{ 2 } }{ 5\omega -3 } = \ ?$

Note: 1, $\omega$ and $\omega ^{ 2 }$ are the cubic roots of the unity.

Note: if it gives you more than one answer, add them together.

The answer is -1.00.

1 solution

Chew-Seong Cheong
Jun 17, 2015

$\omega$ is the cubic root of unity. Therefore, $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$ .

$\begin{aligned} \frac{7+4\omega}{7\omega^2+4} + \frac{5-3\omega^2}{5\omega-3} & = \frac{7\omega^3 +4\omega}{7\omega^2+4} + \frac{5\omega^3 -3\omega^2}{5\omega-3} \\ & = \omega + \omega^2 \\ & = \boxed{-1} \end{aligned}$

Moderator note:

Sometimes we just have to hunt down the coefficients carefully.

A K you have to mention that $\omega$ is the cubic root of unity in the question.

Chew-Seong Cheong - 5 years, 12 months ago

Wow,nice and short soln..good

Akhil Bansal - 5 years, 11 months ago

Free algebra! part 2

The answer is -1.00.

1 solution

Moderator note:

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