Question

Expand (4a - 2b - 3c)^2 using suitable identities,
note :-
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Answer 1

Answer :-

Given :-

(4a - 2b - 3c)²

Using the identity :-

⟹ (x + y + z)² :

• x² + y² + z² + 2xy + 2yz + 2zx

Since :-

⟹ [4a +(-2b)² + (-3c)²]

⟹ (-4a)² + (-2b)² + (-3c)² + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a)

⟹ 16a² + 4b² + 9c² - 16ab + 12bc - 24ac

Learn more :-

(x + y)² = x² + y² + 2xy

(x - y)² = x² - 2xy + y²

x² - y² = (x + y)(x - y)

(x + y)³ = x³ + y³ + 3xy(x + y)

(x - y)³ = x³ - y³ - 3xy(x - y)

x³ - y³ = (x - y)(x² + xy + y²)

x³ + y³ = (x + y)(x² - xy + y²)

Answer 2

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

$\sf \red {\underline{\underline{Provided:-}}}$

$\to \tt {(4a-2b-3c)^{2}}$

➻The given Equation is in the form of,

$\to \tt {(x+y+z)^{2}}$

$\sf \blue {\underline{\underline{We\: know:-}}}$

$\to \tt {(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz}$

where,

$\mapsto \tt {x=4a}$

$\mapsto \tt {y=-2b}$

$\mapsto \tt {z=-3c}$

➻Supplanting the given values in the formula,

$\to \tt {(4a-2b-3c)^{2}=(4a)^{2}+(-2b)^{2}+(-3c)^{2}+2(4a)(-2b)+2(-2b)(-3c)+2(4a)(-3c)}$

$\to \tt {(4a-2b-3c)^{2}=16^{2}+4b^{2}+9c^{2}-16ab+12bc-24ac}$

➻This can't be expanded further.