If $a + b + c = 1$ , $a^{2} + b^{2} + c^{2}$ = $a^{2} + b^{2} + c^{2} = 21$ , $a b c = 8$ . Find the value of $(1 - a) (1 - b) (1 - c)$ .

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Solution

$(1 - a) (1 - c - b + b c) = 1 - c - b + b c - a + a c + a b = 1 - (a + b + c) + (a b + b c + a c) = 1 - 1 + a b + b c + a c = \frac{2 (a b + b c + a c)}{2} = \frac{(a + b + c)^{2} - (a^{2} + b^{2} + c^{2})}{2} = \frac{1 - 21}{2} = - 10$

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Similar questions

\frac{a^{2} - b^{2} - c^{2}}{(a - b) (a - c)} + \frac{b^{2} - c^{2} - a^{2}}{(b - c) (b - a)} + \frac{c^{2} - a^{2} - b^{2}}{(c - a) (c - b)}

a = 1

b = - 1

c = 0

(a^{2} - b^{2}) + (c^{2} - b^{2}) + (a^{2} - c^{2})

Parallel lines are obtained for the pair of equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ if

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

a + b + c = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

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