If ab + bc + ca = 0,then the value of left(fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $ab + bc + ca = 0$.From this,we have $bc = -ab - ca$.Substituting this into the first term denominator: $a^2 - bc = a^2 - (-ab - ca) = a^2 +

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given $ab + bc + ca = 0$.From this,we have $bc = -ab - ca$.Substituting this into the first term denominator: $a^2 - bc = a^2 - (-ab - ca) = a^2 + ab + ca = a(a + b + c)$.Similarly,for the other denominators:$b^2 - ca = b^2 - (-ab - bc) = b^2 + ab + bc = b(a + b + c)$.$c^2 - ab = c^2 - (-bc - ca) = c^2 + bc + ca = c(a + b + c)$.Now,substitute these into the expression:$\frac{1}{a(a + b + c)} + \frac{1}{b(a + b + c)} + \frac{1}{c(a + b + c)}$$= \frac{1}{a + b + c} \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}ight)$$= \frac{1}{a + b + c} \left(\frac{bc + ac + ab}{abc}ight)$Since $ab + bc + ca = 0$,the numerator of the second fraction is $0$.$= \frac{1}{a + b + c} 	imes \frac{0}{abc} = 0$.

If $ab + bc + ca = 0$,then the value of $\left(\frac{1}{a^2 - bc} + \frac{1}{b^2 - ca} + \frac{1}{c^2 - ab}\right)$ is

Similar Questions

If $N = \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}},$ then what is the value of $N + \frac{1}{N} ?$

In a college,$\frac{1}{5}$ of the girls and $\frac{1}{8}$ of the boys took part in a social camp. What part of the total number of students in the college took part in the camp?

$832.58 - 242.31 = 779.84 - ?$

If the numbers $\frac{3}{5}, \frac{2}{3}, \frac{3}{4}$ are given,then we can say that

If $\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} = 0$,then the value of $\frac{1}{a} + \frac{1}{b}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module