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Question Number 179157 by Agnibhoo98 last updated on 25/Oct/22

If x = a^2 − bc, y = b^2 − ca, z = c^2 − ab then prove that, x^3 + y^3 + z^3 − 3xyz is a perfect square.

$$\mathrm{If}\:{x}\:=\:{a}^{\mathrm{2}} −\:{bc},\:{y}\:=\:{b}^{\mathrm{2}} \:−\:{ca},\:{z}\:=\:{c}^{\mathrm{2}} \:−\:{ab} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}, \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \:+\:{z}^{\mathrm{3}} \:−\:\mathrm{3}{xyz}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$

Answered by som(math1967) last updated on 25/Oct/22

x−y=a^2 −bc−b^2 +ca =(a−b)(a+b) +c(a−b) =(a−b)(a+b+c) y−z=(b−c)(a+b+c) z−x=(c−a)(a+b+c) x^3 +y^3 +z^3 −3xyz =(x+y+z)(x^2 +y^2 +z^2 −xy−yz−zx) =(1/2)(x+y+z){(x−y)^2 +(y−z)^2 +(z−x)^2 } =(a^2 +b^2 +c^2 −ab−bc−ca) ×[(a+b+c)^2 {(a−b)^2 +(b−c)^2 +(c−a)^2 } =(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) ×(1/2)(a+b+c){(a−b)^2 +(b−c)^2 +(c−a)^2 } =(a^3 +b^3 +c^3 −3abc)(a^3 +b^3 +c^3 −3abc) =(a^3 +b^3 +c^3 −3abc)^2 perfect square

$$\:{x}−{y}={a}^{\mathrm{2}} −{bc}−{b}^{\mathrm{2}} +{ca} \\ $$$$\:\:=\left({a}−{b}\right)\left({a}+{b}\right)\:+{c}\left({a}−{b}\right) \\ $$$$=\left({a}−{b}\right)\left({a}+{b}+{c}\right) \\ $$$$\:{y}−{z}=\left({b}−{c}\right)\left({a}+{b}+{c}\right) \\ $$$${z}−{x}=\left({c}−{a}\right)\left({a}+{b}+{c}\right) \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}{xyz} \\ $$$$=\left({x}+{y}+{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}+{z}\right)\left\{\left({x}−{y}\right)^{\mathrm{2}} +\left({y}−{z}\right)^{\mathrm{2}} +\left({z}−{x}\right)^{\mathrm{2}} \right\} \\ $$$$=\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$\:\:×\left[\left({a}+{b}+{c}\right)^{\mathrm{2}} \left\{\left({a}−{b}\right)^{\mathrm{2}} +\left({b}−{c}\right)^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} \right\}\right. \\ $$$$=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$×\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}+{c}\right)\left\{\left({a}−{b}\right)^{\mathrm{2}} +\left({b}−{c}\right)^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} \right\} \\ $$$$=\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}\right)\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}\right) \\ $$$$=\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{perfect}}\:\boldsymbol{{square}} \\ $$

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