Add:
$a (a - b), b (c - a)$ and $c (b - a)$

Open in App

Solution

$a (a - b) + b (c - a) + c (b - a)$
$= a (a - b) - c (a - b) + b (c - a)$
$= (a - b) (a - c) + b (c - a)$
$= (a - b) (a - c) - b (a - c)$
$= (a - c) (a - b - b)$
$= (a - c) (a - 2 b)$

Suggest Corrections

Similar questions

Q. Add:

a + b + a b; b - c + b c

and

c + a + a c .

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

The determinant
$∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ equals to:

(a) abc(b-c)(c-a)(a-b)
(b) (b-c)(c-a)(a-b)
(c) (a+b+c)(b-c)(c-a)(a-b)
(d) None of these

Q. Prove that

∣ ∣ ∣ ∣ ∣ \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. Tick (✓) the correct answer:
a² + bc + ab + ac = ?
(a) (a + b)(a + c)
(b) (a + b)(b + c)
(c) (b + c)(c + a)
(d) a(a + b + c)

Join BYJU'S Learning Program

Add:a (a−b),b (c−a) and c (b−a)

a(a−b)+b(c−a)+c(b−a)=a(a−b)−c(a−b)+b(c−a)=(a−b)(a−c)+b(c−a)=(a−b)(a−c)−b(a−c)=(a−c)(a−b−b)=(a−c)(a−2b)

Add:
$a (a - b), b (c - a)$ and $c (b - a)$

$a (a - b) + b (c - a) + c (b - a)$
$= a (a - b) - c (a - b) + b (c - a)$
$= (a - b) (a - c) + b (c - a)$
$= (a - b) (a - c) - b (a - c)$
$= (a - c) (a - b - b)$
$= (a - c) (a - 2 b)$