The correct option is D a2+2ab+b2 c2 (a+b+c)(a+b c) =a(a+b c)+b(a+b c)+c(a+b c) =a^2 + ab ac + ba + b^2 bc + ca + cb c^2 = (a^2 + 2ab + b^2 c^2) ( ∵ a ...

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Byju's Answer

Standard VIII

Mathematics

Multiplication of Polynomials

a+b+ca+b-c=-

Question

$(a + b + c) (a + b - c)$ = ___.

a²+2ab-c²

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a²+4ab+b²-c²

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a²+2ab+b²

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a²+2ab+b²-c²

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Solution

The correct option is D
a²+2ab+b²-c²

$(a + b + c) (a + b - c)$

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

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∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + a b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3. (b - c) (c - a) (a - b) (a + b + c) (a b + b c + c a)

Q. Show that :

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + c b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3 (b - c) (c - a) (a - b) (a + b + c) (b c + c a + a b)

∣ ∣ ∣ ∣ \begin{matrix} b + c & a - c & a - b b - c & c + a & b - a c - b & c - a & a + b \end{matrix} ∣ ∣ ∣ ∣ =

$(a + b + c) (a - b - c)$ = ______

(a+b+c)(a+b-c) = ___.

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(a+b+c)(a+b−c) = ___.

The correct option is D a2+2ab+b2-c2 (a+b+c)(a+b−c) =a(a+b−c)+b(a+b−c)+c(a+b−c) =a2+ab−ac+ba+b2−bc+ca+cb−c2 =(a2+2ab+b2−c2) ( ∵ac=ca and bc=cb )

$(a + b + c) (a + b - c)$ = ___.

The correct option is D
a²+2ab+b²-c²

$(a + b + c) (a + b - c)$

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