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Algebraic Identities

State whether...

Question

State whether the statement is True or False.
Expand: $(a + b - c)^{2}$ is equal to $a^{2} + b^{2} + c^{2} + 2 a b - 2 b c - 2 c a$ .

True

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False

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Solution

The correct option is A True
$M u l t i p l y (a + b - c) (a + b - c) = a^{2} + b^{2} + c^{2} + a b - b c - c a + a b - b c - c a = a^{2} + b^{2} + c^{2} + 2 a b - 2 b c - 2 c a$

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Similar questions

(a - b + c)^{2}

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

△ A B C

cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

△ A B C, a : b : c = cos A : cos B : cos C

△ A B C

cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}

cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

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}}

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