# If ab + bc + ca = 0, then find1/a2-bc + 1/b2 – ca + 1/c2- ab

SHAIK AASIF AHAMED
8 years ago
Hello student,
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
13 Points
6 years ago
Take a=b=1 and get the value of c , so you will get c=-1/2.
put the value of a=b=1 and c=-1/2 in the equation on the right hand side , you will get ans as 0.
Mohammed Faizan Makrani
19 Points
4 years ago
+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=0you get value = 0/[(a+b+c)(abc)] = 02+ac+bc))+(1/(a2+ab+ac))+(1/(c2-bc)Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + acyou get :=(1/(a2-ab) + 1/(a2-bc ) + 1/(c2Given ab+bc+ca=0 and asked to find value 1/(a
Ashutosh
13 Points
4 years ago
a+b+c = 0
(a+b)² = (-c)²
a²+b²+2ab = c²
Squaring on both sides, we get
(a²+b²-c²)² = (-2ab)²
a⁴+b⁴+c⁴+2a²b²-2b²c²-2c²a² = 4a²b² [∴(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
a⁴+b⁴+c⁴+2a²b²-4a²b²-2b²c²-2c²a² = 0
a⁴+b⁴+c⁴-2a²b²-2b²c²-2c²a²= 0
a⁴+b⁴+c⁴= 2a²b²+2b²c²+2c²a²
a⁴+b⁴+c⁴= 2(a²b²+b²c²+c²a²)
15 Points
3 years ago
Given ab+ca,-ca=ab+bc,-ab=bc+ca.
Hence,1/a2-bc+1/b2-ca+1/c2-ab
=1/a2+ab+ca +1/b2+ab+bc +1/c2+bc+ca
=1/a(a+b+c)+1/b(a+b+c)+1/c(a+b+c)
=bc+ca+ab/abc(a+b+c)=0/abc(a+b+c)
=0

Ajeet Tiwari
2 years ago
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