If ab + bc + ca = 0, then find

1/a²-bc + 1/b² – ca + 1/c²- ab

Hello student,
Please find the answer to your question below
Given ab+bc+ca=0 and asked to find value 1/(a²-bc ) + 1/(c²-ab) + 1/(a²-bc)
Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac
you get :
=(1/(a²+ab+ac))+(1/(c²+ac+bc))+(1/(a²+ab+ac))
=(1/a(a+b+c))+(1/b(a+b+c))+(1/c(a+b+c))
By taking LCM we get
=(ab + bc + ca)/[(a+b+c)(abc)]
put, ab+bc+ca=0
you get value = 0/[(a+b+c)(abc)] = 0

Hello student
Given ab + bc + ca = 0
=> bc = - ab - ca = -a (b+c)
=> a² - bc = a (a - b - c)

similarly, b² - ca = b (b - c - a) and c² - ab = c (c - a - b)

==========
1/(a²-bc) + 1/(b²-ca) + 1/(c² -ab)
= [(b²-ca)(c²-ab) + (c²-ab)(a²-bc) + (a²-bc)(b²-ca) ] / [(a²-bc)(b²-ca)(c²-ab)]

we simplify the numerator.
= [ b²c² - ac³ - a b³ +a²bc +a²c² - ba³ - bc³ + ab²c + a²b² - cb³ -ca³ + abc² ]
= [ b²(c²-ab+ac+a²-bc) + c²(-ac+a²-bc+ab) + a² (bc-ab-ca)

we use the given identity.

= [ b² (c² +a²+2 ac) + c² (a² +2ab) + a²(2bc) ]
= [ b² (c + a)² + a² c² + 2a²bc + 2abc² ]
= [ { bc + ba}² + a²c² + 2ac (ab + bc) ]

we use the given identity again.

= [ {-ac}² + a² c² + 2 ac (-ac) ]
= 0

so the answer is 0.


Hope it helps
Thankyou