Just Ordinary Cubing

Algebra Level 2

For complex number J J , where J ≠ − 1 J ≠ -1 and J 2 = J − 1 J^{2} = J - 1 , evaluate J 3 J^{3} .


The answer is -1.

Ir J by IR J , 21, Philippines

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2 solutions

Denis Kartachov
Aug 22, 2018

From

J 3 + 1 = ( J + 1 ) ( J 2 − J + 1 ) J^3 + 1 = (J+1)(J^2-J+1)

We are given J 2 − J + 1 = 0 J^2 - J +1 = 0 so:

J 3 + 1 = 0 ⟹ J 3 = − 1 J^3 + 1 = 0 \Longrightarrow J^3 = -1

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
Aug 28, 2018

J 2 = J − 1 Given J 3 = J 2 − J Multiply both sides by J . = J − 1 − J = − 1 \begin{aligned} J^2 & = J - 1 & \small \color{#3D99F6} \text{Given} \\ J^3 & = {\color{#3D99F6}J^2} - J & \small \color{#3D99F6} \text{Multiply both sides by }J. \\ & = {\color{#3D99F6}J-1} - J \\ & = \boxed{-1} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@Ivan Richmond Jumawan , I think it should be J ≠ 1 J \ne 1 instead of J ≠ − 1 J \ne -1 so that J 2 = J − 1 ≠ 0 J^2 = J-1 \ne 0 and J ≠ 0 J \ne 0 .

Chew-Seong Cheong - 2 years, 9 months ago

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