Verify the validity of the identity
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
for the following given values.
1. $a = 2$ $b = 4$ $c = 6$
2. $a = 5$ $b = - 2$ $c = 3$
3. $a = - 2$ $b = 3$ $c = - 4$

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Solution

i) For $a = 2, b = 4, c = 6$ we have
$(a + b + c)^{2} = (12)^{2} = 144$ and
$a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
$= 4 + 16 + 36 + 16 + 48 + 24$
$= 144$
ii) For $a = 5, b = - 2, c = 3$ we have
$(a + b + c)^{2} = (6)^{2} = 36$ and
$a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
$= 25 + 4 + 9 - 20 - 12 + 30$
$= 36$
iii) For $a = - 2, b = 3, c = - 4$ we have
$(a + b + c)^{2} = (- 3)^{2} = 9$ and
$a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
$= 4 + 9 + 16 - 12 - 24 + 16$
$= 9$

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Similar questions

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

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, 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}}

{(a + b)}^{2} = a^{2} + 2 a b + b^{2}

a = 2, b = 4

A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

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