if a+b+c =0 then 1/2( a^2/bc + b^2/ac + c^2 /ab)
Answers
Step-by-step explanation:
Value of \frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}bca2+cab2+abc2 = 33
Explanation:
Given a+b+c=0 ----(1)
we know the algebraic identity:
\begin{gathered}\boxed { a^{3}+b^{3}+c^{3}\\=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)+3abc}\end{gathered}
Value of \frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}bca2+cab2+abc2
= \frac{a^{3}}{abc}+\frac{b^{3}}{abc}+\frac{c^{3}}{abc}abca3+abcb3+abcc3
= \frac{a^{3}+b^{3}+c^{3}}{abc}abca3+b3+c3
= \frac{(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)+3abc}{abc}abc(a+b+c)(a2+b2+c2−ab−bc−ca)+3abc
= \frac{0\times(a^{2}+b^{2}+c^{2}-ab-bc-ca)+3abc}{abc}abc0×(a2+b2+c2−ab−bc−ca)+3abc
/* from (1) */
= \frac{3abc}{abc}abc3abc
= 33
Therefore,
Value of \frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}bca2+cab2+abc2 = 33