If ab - - ca - 0- then the value of dfrac-1-a-2- - bc- - dfrac-1-b-2- - ca- - dfrac-1-c-2- - ab- will be-a - b - c-10ab

Question

Toppr Admin · Accepted Answer

Ab-bc-ca-0-x2212-x2212-x2212-x2212-1-1a2-x2212-bc-1b2-x2212-ca-1c2-x2212-ab-1a2-ab-ca-1b2-ab-bc-1c2-aa-ca -From -1-1a-1a-b-c-1b-1a-b-c-1c-1a-b-c-1a-b-c-1a-1b-1c-ab-bc-ca-a-b-c-abc-0-xA0- -xA0-xA0-From -1-

If ab+bc+ca=0, then the value of 1a2−bc+1b2−ca+1c2−ab will be:a+b+c−10ab

ab+bc+ca=0−−−−(1)=1a2−bc+1b2−ca+1c2−ab=1a2+ab+ca+1b2+ab+bc+1c2+aa+ca (From (1)=1a(1a+b+c)+1b(1a+b+c)+1c(1a+b+c)=1a+b+c(1a+1b+1c)=ab+bc+ca(a+b+c)(abc)=0 From (1)

If $= \frac{1}{a^{2} + a b + c a} + \frac{1}{b^{2} + a b + b c} + \frac{1}{c^{2} + a a + c a}$ then the value of $\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b}$ will be:
$a + b + c$
$- 1$
$0$
$a b$