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Expansion of (x+y)^3

Expansion of ...

Question

Expansion of $(2 a + 3 b + 4 c)^{2}$ is _____.

4 a^{2} + 9 b^{2} + 16 c^{2} + 12 a b + 24 b c + 16 c a

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

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

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

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Solution

The correct option is A $4 a^{2} + 9 b^{2} + 16 c^{2} + 12 a b + 24 b c + 16 c a$
We know,

$(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$ $[I d e n t i t y]$

Here,
$x = 2 a$
$y = 3 b$
$z = 4 c$

Putting Values in Identity, We get :

$\Rightarrow$ $(2 a + 3 b + 4 c)^{2} = (2 a)^{2} + (3 b)^{2} + (4 c)^{2} + 2 (2 a) (3 b) + 2 (3 b) (4 c) + 2 (4 c) (2 a)$
$= 4 a^{2} + 9 b^{2} + 16 c^{2} + 12 a b + 24 b c + 16 c a$

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Similar questions

If a, b, c are in G.P., prove that:

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

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

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

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

a^{2} + b^{2} + 4 c^{2} + 2 (a b + 2 b c + 2 c a)

(a + b + 2 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}

△ 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}

a, b, c

a^{2} + 4 b^{2} + 16 c^{2} = 48

a b + 4 b c + 2 c a = 24

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

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