Find a2 - 2bc - b2 - c2 - a - b - c a-b-c

Given #160; (a^2 2bc b^2 c^2) ÷ (a b c) (a+b+c) (a b c)(a+b+c)=a^2 ab ac+ba b^2 bc+ca cb c^2=a^2 b^2 c^2 2bc So, (a^2 2bc b^2 c^2 ...

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Addition of Polynomials

Find a2 - 2...

Question

Find $(a^{2} - 2 b c - b^{2} - c^{2}) \div (a - b - c) (a + b + c)$

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Solution

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

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} [{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}]

a² + b² + c² - ab - bc - ca equals:

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