solve the equation: ax+by+cz=0, x+y+z=0, bcx+cay+abz=(b-c)(c-a)(a-b)
Answers
Answer:
Given, ax+by+cz=0→(1)
bcx+cay+abz=0→(2)
xyz+abc(a
3
x+b
3
y+c
3
z)=0→(3)
(1)×bc−(2)×a⇒abcx+b
2
cy+bc
2
z−abcx−ca
2
y−a
2
b
2
=0
c(b
2
−a
2
)y=b(a
2
−c
2
)z
(1)×ac−(2)×b⇒a
2
cx+abcy+ac
2
z−b
2
cx−abcy−ab
2
z=0
c(a
2
−b
2
)x=a(b
2
−c
2
)z
(1)×ab−(2)×c⇒a
2
bx+ab
2
y+abcz−bc
2
x−ac
2
y−abcz=0
b(a
2
−c
2
)x=a(c
2
−b
2
)y
Let c(b
2
−a
2
)y=b(a
2
−c
2
)z=k
y=
c(b
2
−a
2
)
k
,z=
b(a
2
−c
2
)
k
Let c(a
2
−b
2
)x=a(b
2
−c
2
)z=k
1
x=
c(a
2
−b
2
)
k
1
,z=
a(b
2
−c
2
0
k
1
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Given, ax+by+cz=0→(1)
bcx+cay+abz=0→(2)
xyz+abc(a
3
x+b
3
y+c
3
z)=0→(3)
(1)×bc−(2)×a⇒abcx+b
2
cy+bc
2
z−abcx−ca
2
y−a
2
b
2
=0
c(b
2
−a
2
)y=b(a
2
−c
2
)z
(1)×ac−(2)×b⇒a
2
cx+abcy+ac
2
z−b
2
cx−abcy−ab
2
z=0
c(a
2
−b
2
)x=a(b
2
−c
2
)z
(1)×ab−(2)×c⇒a
2
bx+ab
2
y+abcz−bc
2
x−ac
2
y−abcz=0
b(a
2
−c
2
)x=a(c
2
−b
2
)y Let c(b
2
−a
2
)y=b(a
2
−c
2
)z=k
y=
c(b
2
−a
2
)
k
,z=
b(a
2
−c
2
)
k
Let c(a
2
−b
2
)x=a(b
2
−c
2
)z=k
1
x=
c(a
2
−b
2
)
k
1
,z=
a(b
2
−c
2
0
k
1
b(a
2
−c
2
)
k
=
a(b
2
−c
2
)
k
1
k
1
=
b
a
(
a
2
−c
2
b
2
−c
2
)k
x=
bc(a
2
−b
2
)(a
2
−c
2
)
a(c
2
)k
,y=
c(b
2
−a
2
)
k
,z=
b(a
2
−c
2
)
k
xyz=
[bc(b
2
−a
2
)(a
2
−c
2
)]
2
a(c
2
−b
2
)k
3
a
3
x+b
3
y+c
3
z=a
2
(−by−cz)+b
3
y+c
3
z
=−b(a
2
−b
2
)y+c(c
2
−a
2
)z
c
kb
−
b
kc
(3)⇒xyz+abc(a
3
x+b
3
y+c
3
z)=0
(bc(c
2
−a
2
)(a
2
−b
2
))
2
a(c
2
−b
2
)k
3
+kab
2
−kac
2
=0
a(c
2
−b
2
)k[
(bc(c
2
−a
2
)(a
2
−b
2
))
2
k
2
−1]=0
∴k=0 or k=±bc(c
2
−a
2
)(a
2
−b
2
)
∴ solutions are x=0,y=0,z=0 and
x=a(b
2
−c
2
),y=b(a
2
−c
2
),z=c(a
2
−b
2
)
and x=−a(b
2
−c
2
),y=−b(a
2
−c
2
),z=−c(a
2
−b
2
)