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Question Number 139490 by EnterUsername last updated on 27/Apr/21

If w≠1 is a cube root of unity, x=a+b, y=aw+bw^2 and z=aw^2 +bw, then x^3 +y^3 +z^3 =?

$$\mathrm{If}\:{w}\neq\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity},\:\mathrm{x}={a}+{b},\:\mathrm{y}={aw}+{bw}^{\mathrm{2}} \\ $$$$\mathrm{and}\:{z}={aw}^{\mathrm{2}} +{bw},\:\mathrm{then}\:\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =? \\ $$

Answered by Rasheed.Sindhi last updated on 27/Apr/21

x^3 +y^3 +z^3 =(a+b)^3 +(aw+bw^2 )^3 +(aw^2 +bw)^3 =a^3 +b^3 +3ab(a+b) +a^3 w^3 +b^3 (w^3 )^2 +3abw^3 (aw+bw^2 ) +a^3 (w^3 )^2 +b^3 w^3 +3abw^3 (aw^2 +bw) =a^3 +b^3 +3ab(a+b) +a^3 (1)+b^3 (1)^2 +3ab(1)(aw+bw^2 ) +a^3 (1)^2 +b^3 (1)+3ab(1)(aw^2 +bw) =3a^3 +3b^3 +3ab(a+aw+aw^2 +b+bw+bw^2 ) =3a^3 +3b^3 +3ab{a(1+w+w^2 )+b(1+w+w^2 )} =3a^3 +3b^3 +3ab{a(0)+b(0)} =3(a^3 +b^3 )

$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\left({a}+{b}\right)^{\mathrm{3}} +\left({aw}+{bw}^{\mathrm{2}} \right)^{\mathrm{3}} +\left({aw}^{\mathrm{2}} +{bw}\right)^{\mathrm{3}} \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+{a}^{\mathrm{3}} {w}^{\mathrm{3}} +{b}^{\mathrm{3}} \left({w}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{3}{abw}^{\mathrm{3}} \left({aw}+{bw}^{\mathrm{2}} \right) \\ $$$$\:\:+{a}^{\mathrm{3}} \left({w}^{\mathrm{3}} \right)^{\mathrm{2}} +{b}^{\mathrm{3}} {w}^{\mathrm{3}} +\mathrm{3}{abw}^{\mathrm{3}} \left({aw}^{\mathrm{2}} +{bw}\right) \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$$\:\:\:\:\:\:+{a}^{\mathrm{3}} \left(\mathrm{1}\right)+{b}^{\mathrm{3}} \left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}{ab}\left(\mathrm{1}\right)\left({aw}+{bw}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:+{a}^{\mathrm{3}} \left(\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{3}} \left(\mathrm{1}\right)+\mathrm{3}{ab}\left(\mathrm{1}\right)\left({aw}^{\mathrm{2}} +{bw}\right) \\ $$$$=\mathrm{3}{a}^{\mathrm{3}} +\mathrm{3}{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{aw}+{aw}^{\mathrm{2}} +{b}+{bw}+{bw}^{\mathrm{2}} \right) \\ $$$$=\mathrm{3}{a}^{\mathrm{3}} +\mathrm{3}{b}^{\mathrm{3}} +\mathrm{3}{ab}\left\{{a}\left(\mathrm{1}+{w}+{w}^{\mathrm{2}} \right)+{b}\left(\mathrm{1}+{w}+{w}^{\mathrm{2}} \right)\right\} \\ $$$$=\mathrm{3}{a}^{\mathrm{3}} +\mathrm{3}{b}^{\mathrm{3}} +\mathrm{3}{ab}\left\{{a}\left(\mathrm{0}\right)+{b}\left(\mathrm{0}\right)\right\} \\ $$$$=\mathrm{3}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right) \\ $$$$ \\ $$

Commented by EnterUsername last updated on 27/Apr/21

Thanks Sir

$${Thanks}\:{Sir} \\ $$

Answered by Ar Brandon last updated on 27/Apr/21

x^3 +y^3 +z^3 −3xyz=(x+y+z)(x+yw+zw^2 )(x+yw^2 +zw) x^3 +y^3 +z^3 −3(a+b)(aw+bw^2 )(aw^2 +bw)=0 x^3 +y^3 +z^3 =3(a+b)(a^2 w^3 +b^2 w^3 +ab(w^2 +w^4 ) x^3 +y^3 +z^3 =3(a+b)(a^2 −ab+b^2 )=3(a^3 +b^3 )

$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3xy}{z}=\left(\mathrm{x}+\mathrm{y}+{z}\right)\left(\mathrm{x}+\mathrm{y}{w}+{zw}^{\mathrm{2}} \right)\left(\mathrm{x}+\mathrm{y}{w}^{\mathrm{2}} +{zw}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}\left({a}+{b}\right)\left({aw}+{bw}^{\mathrm{2}} \right)\left({aw}^{\mathrm{2}} +{bw}\right)=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{3}\left({a}+{b}\right)\left({a}^{\mathrm{2}} {w}^{\mathrm{3}} +{b}^{\mathrm{2}} {w}^{\mathrm{3}} +{ab}\left({w}^{\mathrm{2}} +{w}^{\mathrm{4}} \right)\right. \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{3}\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)=\mathrm{3}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right) \\ $$

Commented by Ar Brandon last updated on 27/Apr/21

x+y+z=a+b+(w+w^2 )a+(w+w^2 )b =a+b−a−b=0 (w^2 +w=−1), w^4 =w

$$\mathrm{x}+\mathrm{y}+{z}={a}+{b}+\left({w}+{w}^{\mathrm{2}} \right){a}+\left({w}+{w}^{\mathrm{2}} \right){b} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}+{b}−{a}−{b}=\mathrm{0}\:\:\left({w}^{\mathrm{2}} +{w}=−\mathrm{1}\right),\:{w}^{\mathrm{4}} ={w} \\ $$

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