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Question Number 192875 by York12 last updated on 30/May/23
x^2 −yz=a^n y^2 −zx=b^n z^2 −xy=c^n find (x , y , z) in terms of (a , b , c)
$${x}^{\mathrm{2}} −{yz}={a}^{{n}} \\ $$$${y}^{\mathrm{2}} −{zx}={b}^{{n}} \\ $$$${z}^{\mathrm{2}} −{xy}={c}^{{n}} \\ $$$${find}\:\left({x}\:,\:{y}\:,\:{z}\right)\:{in}\:{terms}\:{of}\:\left({a}\:,\:{b}\:,\:{c}\right) \\ $$
Answered by Frix last updated on 30/May/23
My path: 1. y=px∧z=qx 2. p=u+v∧q=u−v It′s then possible to solve the system. I got u=((b^(2n) +c^(2n) −a^n (b^n +c^n ))/(2(a^(2n) −b^n c^n )))∧v=(((a^n +b^n +c^n )(c^n −b^n ))/(2(a^(2n) −b^n c^n ))) p=((b^(2n) −a^n c^n )/(a^(2n) −b^n c^n ))∧q=((c^(2n) −a^n b^n )/(a^(2n) −b^n c^n )) x=±((a^(2n) −b^n c^n )/( (√(a^(3n) +b^(3n) +c^(3n) −3a^n b^n c^n )))) y=±((b^(2n) −a^n c^n )/( (√(a^(3n) +b^(3n) +c^(3n) −3a^n b^n c^n )))) z=±((c^(2n) −a^n b^n )/( (√(a^(3n) +b^(3n) +c^(3n) −3a^n b^n c^n )))) I′m too lazy to type the complete workings but everybody should be able to do it if needed.
$$\mathrm{My}\:\mathrm{path}: \\ $$$$\mathrm{1}.\:{y}={px}\wedge{z}={qx} \\ $$$$\mathrm{2}.\:{p}={u}+{v}\wedge{q}={u}−{v} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{then}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{system}. \\ $$$$\mathrm{I}\:\mathrm{got} \\ $$$${u}=\frac{{b}^{\mathrm{2}{n}} +{c}^{\mathrm{2}{n}} −{a}^{{n}} \left({b}^{{n}} +{c}^{{n}} \right)}{\mathrm{2}\left({a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} \right)}\wedge{v}=\frac{\left({a}^{{n}} +{b}^{{n}} +{c}^{{n}} \right)\left({c}^{{n}} −{b}^{{n}} \right)}{\mathrm{2}\left({a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} \right)} \\ $$$${p}=\frac{{b}^{\mathrm{2}{n}} −{a}^{{n}} {c}^{{n}} }{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} }\wedge{q}=\frac{{c}^{\mathrm{2}{n}} −{a}^{{n}} {b}^{{n}} }{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} } \\ $$$${x}=\pm\frac{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} }{\:\sqrt{{a}^{\mathrm{3}{n}} +{b}^{\mathrm{3}{n}} +{c}^{\mathrm{3}{n}} −\mathrm{3}{a}^{{n}} {b}^{{n}} {c}^{{n}} }} \\ $$$${y}=\pm\frac{{b}^{\mathrm{2}{n}} −{a}^{{n}} {c}^{{n}} }{\:\sqrt{{a}^{\mathrm{3}{n}} +{b}^{\mathrm{3}{n}} +{c}^{\mathrm{3}{n}} −\mathrm{3}{a}^{{n}} {b}^{{n}} {c}^{{n}} }} \\ $$$${z}=\pm\frac{{c}^{\mathrm{2}{n}} −{a}^{{n}} {b}^{{n}} }{\:\sqrt{{a}^{\mathrm{3}{n}} +{b}^{\mathrm{3}{n}} +{c}^{\mathrm{3}{n}} −\mathrm{3}{a}^{{n}} {b}^{{n}} {c}^{{n}} }} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{too}\:\mathrm{lazy}\:\mathrm{to}\:\mathrm{type}\:\mathrm{the}\:\mathrm{complete}\:\mathrm{workings} \\ $$$$\mathrm{but}\:\mathrm{everybody}\:\mathrm{should}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{do}\:\mathrm{it}\:\mathrm{if} \\ $$$$\mathrm{needed}. \\ $$
Commented by York12 last updated on 30/May/23
thanks so much sir that was more than enough
$${thanks}\:{so}\:{much}\:{sir}\:{that}\:{was}\:{more}\:{than} \\ $$$${enough} \\ $$
Commented by York12 last updated on 30/May/23
you have exploited that all of them are homogeneous equations of the second degree
$${you}\:{have}\:{exploited}\:{that}\:{all}\:{of}\:{them}\:{are}\: \\ $$$${homogeneous}\:{equations}\:{of}\:{the}\:{second}\:{degree} \\ $$
Answered by MM42 last updated on 30/May/23
1)x^2 −yz=a^n →^(×y) x^2 y−y^2 z=a^n y (i) 2)y^2 −zx=b^n →^(×z) y^2 z−z^2 x=b^n z (ii) 3)z^2 −xy=c^n →^(×x) xz^2 −x^2 y=c^n x (iii) (i)+(ii)+(iii)→c^n x+a^n y+b^n z=0 (iv) if 1)→^(×z) & 2)→^(×x) & 3)→^(×y) and sum of obtined relation→ b^n x+c^n y+a^n z=0 (v) a^n ×(iv)−b^n ×(v)→(a^n c^n −b^(2n) )x+(a^(2n) −b^n c^n )y=0 ⇒x=((a^(2n) −b^n c^n )/(b^(2n) −a^n c^n )) y (vi) c^n ×(iv)−a^n ×(v)→(c^(2n) −a^n b^n )x+(b^n c^n −a^(2n) )z=0 ⇒x=((a^(2n) −b^n c^n )/(a^n b^n −c^(2n) )) z (vii) (vi) & (vii) ⇒ (x/(a^(2n) −b^n c^n ))=(y/(b^(2n) −a^n c^n ))=(z/(a^n b^n −c^(2n) ))
$$\left.\mathrm{1}\right){x}^{\mathrm{2}} −{yz}={a}^{{n}} \:\overset{×{y}} {\rightarrow}\:{x}^{\mathrm{2}} {y}−{y}^{\mathrm{2}} {z}={a}^{{n}} {y}\:\:\left({i}\right) \\ $$$$\left.\mathrm{2}\right){y}^{\mathrm{2}} −{zx}={b}^{{n}} \:\overset{×{z}} {\rightarrow}\:{y}^{\mathrm{2}} {z}−{z}^{\mathrm{2}} {x}={b}^{{n}} {z}\:\:\:\left({ii}\right) \\ $$$$\left.\mathrm{3}\right){z}^{\mathrm{2}} −{xy}={c}^{{n}} \:\overset{×{x}} {\rightarrow}\:{xz}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}={c}^{{n}} {x}\:\:\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right)\rightarrow{c}^{{n}} {x}+{a}^{{n}} {y}+{b}^{{n}} {z}=\mathrm{0}\:\:\:\left({iv}\right) \\ $$$$\left.{i}\left.{f}\left.\:\:\mathrm{1}\right)\overset{×{z}} {\rightarrow}\:\:\&\:\mathrm{2}\right)\overset{×{x}} {\rightarrow}\:\&\:\mathrm{3}\right)\overset{×{y}} {\rightarrow}\:\:{and}\:\:{sum}\:{of}\:{obtined}\:{relation}\rightarrow\:{b}^{{n}} {x}+{c}^{{n}} {y}+{a}^{{n}} {z}=\mathrm{0}\:\:\left({v}\right) \\ $$$${a}^{{n}} ×\left({iv}\right)−{b}^{{n}} ×\left({v}\right)\rightarrow\left({a}^{{n}} {c}^{{n}} −{b}^{\mathrm{2}{n}} \right){x}+\left({a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} \right){y}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} }{{b}^{\mathrm{2}{n}} −{a}^{{n}} {c}^{{n}} }\:{y}\:\:\:\left({vi}\right) \\ $$$${c}^{{n}} ×\left({iv}\right)−{a}^{{n}} ×\left({v}\right)\rightarrow\left({c}^{\mathrm{2}{n}} −{a}^{{n}} {b}^{{n}} \right){x}+\left({b}^{{n}} {c}^{{n}} −{a}^{\mathrm{2}{n}} \right){z}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} }{{a}^{{n}} {b}^{{n}} −{c}^{\mathrm{2}{n}} }\:{z}\:\:\:\:\left({vii}\right) \\ $$$$\left({vi}\right)\:\&\:\left({vii}\right)\:\Rightarrow\:\frac{{x}}{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} }=\frac{{y}}{{b}^{\mathrm{2}{n}} −{a}^{{n}} {c}^{{n}} }=\frac{{z}}{{a}^{{n}} {b}^{{n}} −{c}^{\mathrm{2}{n}} } \\ $$
Commented by York12 last updated on 30/May/23
good job sir
$${good}\:{job}\:{sir}\: \\ $$
Commented by Frix last updated on 30/May/23
But the job is not done yet...
$$\mathrm{But}\:\mathrm{the}\:\mathrm{job}\:\mathrm{is}\:\mathrm{not}\:\mathrm{done}\:\mathrm{yet}… \\ $$
Commented by York12 last updated on 30/May/23
the rest is trivial sir
$${the}\:{rest}\:{is}\:{trivial}\:{sir} \\ $$$$ \\ $$
Commented by York12 last updated on 30/May/23
1)x^2 −yz=a^n →^(×y) x^2 y−y^2 z=a^n y (i) 2)y^2 −zx=b^n →^(×z) y^2 z−z^2 x=b^n z (ii) 3)z^2 −xy=c^n →^(×x) xz^2 −x^2 y=c^n x (iii) (i)+(ii)+(iii)→c^n x+a^n y+b^n z=0 (iv) if 1)→^(×z) & 2)→^(×x) & 3)→^(×y) and sum of obtined relation→ b^n x+c^n y+a^n z=0 (v) a^n ×(iv)−b^n ×(v)→(a^n c^n −b^(2n) )x+(a^(2n) −b^n c^n )y=0 ⇒x=((a^(2n) −b^n c^n )/(b^(2n) −a^n c^n )) y (vi) c^n ×(iv)−a^n ×(v)→(c^(2n) −a^n b^n )x+(b^n c^n −a^(2n) )z=0 ⇒x=((a^(2n) −b^n c^n )/(a^n b^n −c^(2n) )) z (vii) (vi) & (vii) ⇒ (x/(a^(2n) −b^n c^n ))=(y/(b^(2n) −a^n c^n ))=(z/(a^n b^n −c^(2n) ))=λ by subtituting we get λ^2 (a^(3n) +b^(3n) +c^(3n) −3a^n b^n c^n )=1 λ=((+_− 1)/( (√(a^(3n) +b^(3n) +c^(3n) −3a^n b^n c^n )))) ∴ x= ((+_− (a^(2n) −b^n c^n ))/( (√(a^(3n) +b^(3n) +c^(3n) −3a^n b^n c^n )))) , y=((+_− (b^(2n) −c^n a^n ))/( (√(a^(3n) +b^(3n) +c^(3n) −3a^n b^n c^n )))) z= ((+_− (c^(2n) −a^n b^n ))/( (√(a^(3n) +b^(3n) +c^(3n) −3a^n b^n c^n ))))
$$ \\ $$$$\left.\mathrm{1}\right){x}^{\mathrm{2}} −{yz}={a}^{{n}} \:\overset{×{y}} {\rightarrow}\:{x}^{\mathrm{2}} {y}−{y}^{\mathrm{2}} {z}={a}^{{n}} {y}\:\:\left({i}\right) \\ $$$$\left.\mathrm{2}\right){y}^{\mathrm{2}} −{zx}={b}^{{n}} \:\overset{×{z}} {\rightarrow}\:{y}^{\mathrm{2}} {z}−{z}^{\mathrm{2}} {x}={b}^{{n}} {z}\:\:\:\left({ii}\right) \\ $$$$\left.\mathrm{3}\right){z}^{\mathrm{2}} −{xy}={c}^{{n}} \:\overset{×{x}} {\rightarrow}\:{xz}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}={c}^{{n}} {x}\:\:\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right)\rightarrow{c}^{{n}} {x}+{a}^{{n}} {y}+{b}^{{n}} {z}=\mathrm{0}\:\:\:\left({iv}\right) \\ $$$$\left.{i}\left.{f}\left.\:\:\mathrm{1}\right)\overset{×{z}} {\rightarrow}\:\:\&\:\mathrm{2}\right)\overset{×{x}} {\rightarrow}\:\&\:\mathrm{3}\right)\overset{×{y}} {\rightarrow}\:\:{and}\:\:{sum}\:{of}\:{obtined}\:{relation}\rightarrow\:{b}^{{n}} {x}+{c}^{{n}} {y}+{a}^{{n}} {z}=\mathrm{0}\:\:\left({v}\right) \\ $$$${a}^{{n}} ×\left({iv}\right)−{b}^{{n}} ×\left({v}\right)\rightarrow\left({a}^{{n}} {c}^{{n}} −{b}^{\mathrm{2}{n}} \right){x}+\left({a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} \right){y}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} }{{b}^{\mathrm{2}{n}} −{a}^{{n}} {c}^{{n}} }\:{y}\:\:\:\left({vi}\right) \\ $$$${c}^{{n}} ×\left({iv}\right)−{a}^{{n}} ×\left({v}\right)\rightarrow\left({c}^{\mathrm{2}{n}} −{a}^{{n}} {b}^{{n}} \right){x}+\left({b}^{{n}} {c}^{{n}} −{a}^{\mathrm{2}{n}} \right){z}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} }{{a}^{{n}} {b}^{{n}} −{c}^{\mathrm{2}{n}} }\:{z}\:\:\:\:\left({vii}\right) \\ $$$$\left({vi}\right)\:\&\:\left({vii}\right)\:\Rightarrow\:\frac{{x}}{{a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} }=\frac{{y}}{{b}^{\mathrm{2}{n}} −{a}^{{n}} {c}^{{n}} }=\frac{{z}}{{a}^{{n}} {b}^{{n}} −{c}^{\mathrm{2}{n}} }=\lambda \\ $$$${by}\:{subtituting} \\ $$$${we}\:{get}\: \\ $$$$\lambda^{\mathrm{2}} \left({a}^{\mathrm{3}{n}} +{b}^{\mathrm{3}{n}} +{c}^{\mathrm{3}{n}} −\mathrm{3}{a}^{{n}} {b}^{{n}} {c}^{{n}} \right)=\mathrm{1} \\ $$$$\lambda=\frac{\underset{−} {+}\mathrm{1}}{\:\sqrt{{a}^{\mathrm{3}{n}} +{b}^{\mathrm{3}{n}} +{c}^{\mathrm{3}{n}} −\mathrm{3}{a}^{{n}} {b}^{{n}} {c}^{{n}} }} \\ $$$$\therefore\:{x}=\:\frac{\underset{−} {+}\left({a}^{\mathrm{2}{n}} −{b}^{{n}} {c}^{{n}} \right)}{\:\sqrt{{a}^{\mathrm{3}{n}} +{b}^{\mathrm{3}{n}} +{c}^{\mathrm{3}{n}} −\mathrm{3}{a}^{{n}} {b}^{{n}} {c}^{{n}} }}\:\:\:,\:{y}=\frac{\underset{−} {+}\left({b}^{\mathrm{2}{n}} −{c}^{{n}} {a}^{{n}} \right)}{\:\sqrt{{a}^{\mathrm{3}{n}} +{b}^{\mathrm{3}{n}} +{c}^{\mathrm{3}{n}} −\mathrm{3}{a}^{{n}} {b}^{{n}} {c}^{{n}} }} \\ $$$${z}=\:\frac{\underset{−} {+}\left({c}^{\mathrm{2}{n}} −{a}^{{n}} {b}^{{n}} \right)}{\:\sqrt{{a}^{\mathrm{3}{n}} +{b}^{\mathrm{3}{n}} +{c}^{\mathrm{3}{n}} −\mathrm{3}{a}^{{n}} {b}^{{n}} {c}^{{n}} }} \\ $$

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