Question

if a =4 b = 5 c = 6, then find the value of [ (ab+bc+ca-a2-b2-c2)/(3abc-a3 -b3 - c3)

Answer 1

$\textbf{ \huge{ Hello Mate! }}$

$= \frac{ab + bc + ca - {a}^{2} - {b}^{2} - {c}^{2} }{3abc - {a}^{3} - {b}^{3} - {c}^{3} } \\ \\ = \frac{ - ( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - c)}{ - ( {a}^{3} + {b}^{3} + {c}^{3} - 3abc) } \\ \\ = \frac{({a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) }{(a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)} \\ \\ = \frac{1}{a + b + c} = \frac{1}{4 + 5 + 6} \\ \\ = \frac{1}{15}$

$\textsf{\red{ Hence answer is 1 / 15 }}$

$\textbf{ Have great future ahead! }$