If a+b+c=0, then find the value of frac{1}{b^ | vedclass

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

Question

(A) Given $a+b+c=0.$ We need to find the value of $\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}+\frac{1}{a^{2}+b^{2}-c^{2}}.$ Since $a+b+c=

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given $a+b+c=0.$ We need to find the value of $\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}+\frac{1}{a^{2}+b^{2}-c^{2}}.$ Since $a+b+c=0,$ we have $a+b=-c.$ Squaring both sides,$(a+b)^{2}=(-c)^{2},$ which gives $a^{2}+b^{2}+2ab=c^{2}.$ Rearranging,$b^{2}+c^{2}-a^{2} = b^{2}+(a^{2}+2ab+b^{2})-a^{2} = 2b^{2}+2ab = 2b(a+b).$ Since $a+b=-c,$ we get $b^{2}+c^{2}-a^{2} = 2b(-c) = -2bc.$ Similarly,$c^{2}+a^{2}-b^{2} = -2ac$ and $a^{2}+b^{2}-c^{2} = -2ab.$ Substituting these into the expression: $\frac{1}{-2bc} + \frac{1}{-2ac} + \frac{1}{-2ab} = \frac{-a-b-c}{2abc}.$ Since $a+b+c=0,$ the numerator is $0.$ Thus,the value of the expression is $0.$

If $a+b+c=0,$ then find the value of $\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}+\frac{1}{a^{2}+b^{2}-c^{2}}.$

Similar Questions

If the equations $2x^2 + 3x + 5\lambda = 0$ and $x^2 + 2x + 3\lambda = 0$ have a common root,then $\lambda = $

When $x+\frac{1}{x}=3$,find $x^{2}+\frac{1}{x^{2}}$.

If $P(x) = ax^2 + bx + c$ and $Q(x) = -ax^2 + dx + c$ where $ac \neq 0$,then $P(x) \cdot Q(x) = 0$ has at least

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the value of $\alpha\beta^2 + \alpha^2\beta + \alpha\beta$ will be:

Solve the given two equations and select the correct answer from the given options.
$I. 3x^2 - 4x - 32 = 0$
$II. 2y^2 - 17y + 36 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module