(ab+c) +( ba+c ) + (ca+b)= 1 Then (a cube + b cube + c cube) abc =?

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If a +b +c =5 and ab +bc+ca =10,then prove that a cube +b cube+c cube -3abc =-25

If the roots of the equation (csq-ab) xsq-2(asq-bc)x+bsq-ac=0 are equal prove that a=0 /a cube +b cube+c cube=3abc

S = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

A = ⎡ ⎢ ⎣ \begin{matrix} b + c & c + a & b - c c - b & c + b & a - b b - c & a - c & a + b \end{matrix} ⎤ ⎥ ⎦ (a, b, c \neq 0)

^{- 1}

a, b, c

θ = {tan}^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + {tan}^{- 1} \sqrt{\frac{b (a + b + c)}{c a}} + {tan}^{- 1} \sqrt{\frac{c (a + b + c)}{a b}}

tan θ

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