Question

factorise 9(a+b+c)^2-9​

Answer 1

Solution

$9(a + b + c) {}^{2} - 9 \\ = 9((a + b + c) {}^{2} - 1 {}^{2} ) \\ =3 \times 3 (a + b + c - 1)(a + b + c + 1)$

hope this helps you.....

Answer 2

Answer:

9[a²+b²+c²+2ab+2bc+2ca-1]

Step-by-step explanation:

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

substitute it