Question

WHAT IS THE EXPLANATION ​

Answer 1

Question:-

7. If a + b + c = 0, then write the value of$\sf \frac{a {}^{2} }{bc} + \frac{b {}^{2} }{ca} + \frac{c {}^{2} }{ ab}$.

Given:-

• a + b + c = 0.

To Find:-

The value of $\sf \frac{a {}^{2} }{bc} + \frac{b {}^{2} }{ca} + \frac{c {}^{2} }{ab}$.

Solution:-

We know that,

${ \boxed{ {{ \boxed{\sf \large \color{red} a {}^{3} + b {}^{3} + c {}^{3} = (a + b + c)(a {}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca) + 3abc. }}} }}$

If a + b + c = 0 then, a³ + b³ + c³ = 3abc.

$\sf \large = \frac{3{{ \cancel{abc}}}}{{{ \cancel{abc}}}} \\ \\ \\ \sf \large = 3. \: \: \: \: \: \: \:$

Answer:-

${ \boxed{ \sf \large \blue{ \frac{a {}^{2} }{ bc } + \frac{b {}^{2} }{ca} + \frac{c {}^{2} }{ab} = 3. }}}$

Hope you have satisfied. ⚘

Answer 2

${ \underline{ \large{ \sf{Solution : }}}}$

Given Expression,

$\displaystyle \rm{ \frac{a^2}{bc}+ \frac{b^2}{ca} + \frac{ c^2}{ab}}$

When,

• a+b+c=0

Now,

${\implies\displaystyle \rm{ \frac{a^3+b^3+c^3}{abc }}}$

$\therefore \boxed{\displaystyle \rm{a + b + c = 0}}$

Then,

${ \implies\displaystyle \rm{a^3+b^3+c^3=3abc}}$

${\implies\displaystyle \rm{ \frac{a^3+b^3+c^3}{abc }}}$

$\implies\displaystyle \rm{\cancel \frac{3abc}{abc} }$

$\implies\displaystyle \rm{ = 03 }$

Therefore,

${ \color{pink}{\displaystyle \rm{ \frac{a^2}{bc}+ \frac{b^2}{ca} + \frac{ c^2}{ab} = 3}}}$

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Additional Information

$\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{ \:Algrbraic\:Identities }}}} \\ \\ \sf{(a + b)² = a² + 2ab + b² }\\ \\ \sf{(a - b)² = a² - 2ab + b²} \\ \\\sf{a²-b² = (a + b)(a - b) } \\ \\ \sf{(a + b + c)² = a² + b² + c² + 2(ab + bc +ac) }\\ \\ \sf{ (a + b)³ = a³ + 3ab(a + b) + b³ }\\ \\ \sf{(a - b)³ = a³-3ab(a - b) - b³} \\ \\ \sf{a³ + b {}^{3} = (a + b)(a²- ab + b²)} \\ \\ \sf{ 3 a³-b³ = (a - b)(a² + ab + b²) }\\ \\ \sf{(x + a)(x + b) = x² + (a + b)x +ab }\\ \\ \sf{(x + a)(x - b) = x² + (a - b)x - ab} \\ \\ \sf{(x-a)(x + b) = x² - (a - b)x - ab }\\ \\ \sf{(x-a)(x - b) = x² - (a + b)x+ ab} \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$