Question

if a+b+c = 0, then prove that (b+c)²/3bc + (c+a)²/3ac + (a+b)²/3ab = 1

Answer 1

Answer:

(b+c)²/3bc+(c+a)²/3ac+(a+b)²/3ab

=(b²+2bc+c²)/3bc+(c²+2ac+a²)/3ac+(a²+2ab+b²)/3ab

=(ab²+2abc+ac²+bc²+2abc+a²b+a²c+2abc+b²c)/3abc

={ab(a+b)+bc(b+c)+ac(a+c)+6abc}/3abc

=(-abc-abc-abc+6abc)/3abc [∵, a+b+c=0,∴,a+b=-c,b+c=-a,a+c=-b]

=(6abc-3abc)/3abc

=3abc/3abc

=1 (Proved)

Answer 2

ANSWER:

Given:

• a + b + c = 0

Prove that:

• (b + c)²/3bc + (c + a)²/3ac + (a + b)²/3ab = 1

Proof:

$\text{We are given that,}\\\\:\implies a+b+c=0\\\\\text{So,}\\\\:\implies a+b=-c - - - -(1)\\\\:\implies b+c=-a - - - -(2)\\\\:\implies c+a=-b - - - -(3)\\\\\text{We need to prove that,}\\\\:\longrightarrow\dfrac{(b+c)^2}{3bc}+\dfrac{(c+a)^2}{3ac}+\dfrac{(a+b)^2}{3ab}=1$

$\text{Solving LHS,}\\\\:\implies\dfrac{(b+c)^2}{3bc}+\dfrac{(c+a)^2}{3ac}+\dfrac{(a+b)^2}{3ab}\\\\\text{From (1), (2) \& (3),}\\\\:\implies\dfrac{(-a)^2}{3bc}+\dfrac{(-b)^2}{3ac}+\dfrac{(-c)^2}{3ab}$

$\implies\dfrac{a^2}{3bc}+\dfrac{b^2}{3ac}+\dfrac{c^2}{3ab}\\\\\text{Taking LCM = 3abc,}\\\\:\implies\dfrac{(a^2\times a)+(b^2\times b)+(c^2\times c)}{3abc}$

$:\implies\dfrac{a^3+b^3+c^3}{3abc}\\\\\text{We know that,}\\\\:\hookrightarrow\text{If,}\: x+y+z=0,\:\text{then,}\\\\:\implies x^3+y^3+z^3=3abc$

$\text{So,}\\\\:\implies\dfrac{a^3+b^3+c^3}{3abc}$

$:\implies\dfrac{3abc}{3abc}\!\!\!\!\!\!\!\!\!\!\huge{/}$

$\bf{:\implies 1=RHS}\\\\\text{\bf{HENCE PROVED!!!}}$

Formula Used:

$:\hookrightarrow\text{If,}\:x+y+z=0,\:\text{then,}\\\\:\implies x^3+y^3+z^3=3abc$