﻿ Prove: [(2a-b-c)+(b-c)\sqrt{3}i)^3=[(2b-c-a)+(c-a)\sqrt{3}i]^3

#### Question

Prove: [(2a-b-c)+(b-c)\sqrt{3}i)^3=[(2b-c-a)+(c-a)\sqrt{3}i]^3

Collected in the board: Imaginary number

Steven Zheng posted 1 month ago

#### Answer

Expand brackets and combine the like terms

LHS = [(2a-b-c)+(b-c)\sqrt{3}i)^3

=(2a-b+b\sqrt{3}i-c-c\sqrt{3}i )^3

=[2a+b(-1+\sqrt{3}i) +c(-1-\sqrt{3}i )]^3

=(2a+2b\dfrac{-1+\sqrt{3}i}{2}+2c\dfrac{-1-\sqrt{3}i}{2} )^3

Let ω = \dfrac{-1+\sqrt{3}i}{2} , then \overline{ω} = \dfrac{-1-\sqrt{3}i}{2}

ω and \overline{ω} are a pair of complex conjugate, which have the attribute ω^3=1, which means it yields 1 when raised to the power 3. Than,

LHS =(2a+2bω+2c\overline{ω} )^3

Similarly, Expand brackets and combine the like terms of the right hand side.

RHS = [(2b-c-a)+(c-a)\sqrt{3}i]^3

=(2b-c+c\sqrt{3}i-a-a\sqrt{3} i )^3

=(2b+2cω+2a\overline{ω} )^3

=1\cdot (2b+2cω+2a\overline{ω} )^3

=ω^3 (2b+2cω+2a\overline{ω} )^3

=(2a\overline{ω}ω+2bω+2cω^2 )^3

=(2a+2bω+2c\overline{ω} )^3

Therefore, the expressions on the LHS and RHS are equal.

Steven Zheng posted 1 month ago

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