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Factorization of a Quadratic Trinomial

Question 84 v...

Question

Question 84 (viii)

Simplify

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

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Solution

$(a b - c)^{2} + 2 a b c$
$= (a b)^{2} + c^{2} - 2 a b c + 2 a b c$ [using the identity, $(a - b)^{2} = a^{2} + b^{2} - 2 a b$ ]
$= a^{2} b^{2} + c^{2}$

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Simplify :

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

P_{1} : c y + b z = x

P_{2} : a z + c x = y

P_{3} : b x + a y = z .

P_{1}, P_{2}

P_{3}

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

| A_{3 \times 3} | = 3, | B_{3 \times 3} | = - 1

| C_{2 \times 2} | = + 2

| 2 A B C | =

△ A B C

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

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