Simplify: $(a + b + c)^{2} + (a - b + c)^{2}$

2 [a^{2} + b^{2} + c^{2} + 2 a c]

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2 [a^{2} + b^{2} + c^{2} + a c]

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2 [a^{2} + b^{2} + c^{2} - a c]

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

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Solution

The correct option is A $2 [a^{2} + b^{2} + c^{2} + 2 a c]$
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 x z + 2 y z$

So, the given expression,
${(a + b + c)}^{2} + {(a - b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 a c + 2 b c + a^{2} + (- b)^{2} + c^{2} + 2 a (- b) + 2 a c + 2 (- b) c$

$= 2 a^{2} + 2 b^{2} + 2 c^{2} + 2 a b + 2 a c + 2 b c - 2 a b + 2 a c - 2 b c$

$= 2 a^{2} + 2 b^{2} + 2 c^{2} + 4 a c$

Suggest Corrections

Similar questions

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

Δ A B C

c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

\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 x^{2} + b x + c = 0,

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