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Factorial

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Question

Simplify: $\frac{1}{(a - b) (a - c)} + \frac{1}{(b - c) (b - a)} + \frac{1}{(c - a) (b - c)}$

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Solution

$\frac{1}{(a - b) (a - c)} + \frac{1}{(b - c) (b - a)} + \frac{1}{(c - a) (b - c)}$

$= \frac{- b + c}{- (a - b) (a - c) (b - c)} + \frac{- c + a}{- (a - b) (a - c) (b - c)} + \frac{- b + a}{- (a - b) (a - c) (b - c)}$

$= \frac{- b + c - c + a - b + a}{- (a - b) (a - c) (b - c)}$

$= - \frac{2 a - 2 b}{(a - b) (a - c) (b - c)}$

$= - \frac{2}{(a - c) (b - c)}$

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Q. Simplify the following expression:

\frac{1}{(a - b) (a - c)} + \frac{1}{(b - c) (b - a)} + \frac{1}{(c - a) (c - b)}

Q. Using properties of determinants, prove that:

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c b a & b^{2} + 1 & b c c a & c b & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a^{2} + b^{2} + c^{2} + 1

(1 + a + b) (a + b + a b) - a b

\frac{(x - b) (x - c)}{(a - b) (a - c)} + \frac{(x - c) (x - a)}{(b - c) (b - a)} + \frac{(x - a) (x - b)}{(c - a) (c - b)} = 1

D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c b a & b^{2} + 1 & b c c a & c b & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

D =

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