prove that
[ 1/a² bc b+c
1/b² ca c+a
1/c² ab a+b ] = 0
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0
Answer:
ab+bc+ca=0
bc=−ca−ab
bc=−a(b+c)---eq1
like wise,
ab=−c(a+b)---eq2
& ca=−b(c+a)---eq3
=
a
2
−bc
1
+
b
2
−ca
1
+
c
2
−ab
1
taking the value from eq1, 2&3
=
a
2
+a(b+c)
1
+
b
2
+b(a+c)
1
+
c
2
+c(a+b)
1
=
a(a+b+c)
1
+
b(a+b+c)
1
+
c(a+b+c)
1
=
abc(a+b+c)
ab+bc+ca
As ab+bc+ca=0
So, =0
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