The correct option is D 2bc 2ac b^2 c^2(a b)(b c)+(b c)(c a) = (a× b +a× ( c)+( b)× b+( b)× ( c)+b× c+b× ( a) +( c)× (c) + ( c)× a =ab ac b^2+bc+bc ab c^2 ac ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (x+y+z)^2

a-bb-c+b-cc-a...

Question

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

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

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2 a b + 2 b c + 2 a c

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

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

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Solution

The correct option is D $2 b c - 2 a c - b^{2} - c^{2}$
$(a - b) (b - c) + (b - c) (c - a) = (a \times b + a \times (- c) + (- b) \times b + (- b) \times (- c) + b \times c + b \times (- a) + (- c) \times (c) + (- c) \times a = a b - a c - b^{2} + b c + b c - a b - c^{2} - a c = 2 b c - 2 a c - b^{2} - c^{2}$

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Similar questions

\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - a^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}} =

(a) 3(a + b) ( b+ c) (c + a)

(b) 3(a − b) (b − c) (c − a)

(c) (a − b) (b − c) (c − a)

(d) none of these

Q. The expression (a − b)³ + (b − c)³ + (c −a)³ can be factorized as

(a) (a − b) (b − c) (c −a)

(b) 3(a − b) (b − c) (c −a)

(c) −3(a − b) (b −c) (c − a)

(d) (a + b + c) (a² + b² + c² − ab − bc − ca)

$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a)$

$∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a) (a + b + c)$

Q. Add

a (a - b), b (b - c), c (c - a)

Q. Using the properties of determinants, show that:
(i)

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a)

(ii)

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a) (a + b + c)

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(a−b)(b−c)+(b−c)(c−a)= ________

The correct option is D 2bc−2ac−b2−c2(a−b)(b−c)+(b−c)(c−a)=(a×b+a×(−c)+(−b)×b+(−b)×(−c)+b×c+b×(−a)+(−c)×(c)+(−c)×a=ab−ac−b2+bc+bc−ab−c2−ac=2bc−2ac−b2−c2

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