Question

if a b c are on non zero and a + b + c is equals to zero prove that a square up on BC + b square Upon A C + c square Upon A B is equals to 3

Answer 1

if a+b+c=0

then a3+b3+c3=3abc

a2/bc + b2/ca + c2/ab = 3

LCM of bc , ca , ab = abc

a3 + b3 + c3 / abc = 3

3abc / abc = 3

3 = 3

Answer 2

Answer:

Hence Proved that $\frac { a ^ { 2 } } { b c } + \frac { b ^ { 2 } } { c a } + \frac { c ^ { 2 } } { a b } = 3$

To find:

Prove that a square up on BC + b square Upon AC + c square Upon AB is equals to 3

Solution:

Given : a b c are non-zero and a + b + c is equals to zero  

If a + b + c = 0, then $a^3+b^3+c^3$ will becomes,

$\begin{array} { c } { a ^ { 3 } + b ^ { 3 } + c ^ { 3 } = 3 a b c } \\\\ { \frac { 3 a } { a b c } + \frac { 3 b } { a b c } + \frac { 3 c } { a b c } = 3 } \\\\ { \frac { a ^ { 2 } } { b c } + \frac { b ^ { 2 } } { c a } + \frac { c ^ { 2 } } { a b } = 3 } \end{array}$

Taking LCM of bc , ca , ab i.e. abc

$\begin{array} { c } { \frac { a ^ { 3 } } { a b c } + \frac { b ^ { 3 } } { b c a } + \frac { c ^ { 3 } } { a b c } = 3 } \\\\ { \frac { a ^ { 3 } + b ^ { 3 } + c ^ { 3 } } { a b c } = 3 } \end{array}$

We know that the value of $a^3+b^3+c^3=3abc$

Substituting this in the above expression, we get,

$\begin{array} { c } { \frac { 3 a b c } { a b c } = 3 } \\\\ { 3 = 3 } \end{array}$

Thus proved that,

$\frac { a ^ { 2 } } { b c } + \frac { b ^ { 2 } } { c a } + \frac { c ^ { 2 } } { a b } = 3$