Question

factorise 3yz(a-b-c)-3z(a-b-c)+21z²(a-b-c)² can someone answer this​

Answer 1

(a - b - c)²

= (a - b - c)(a - b - c)

= a² - ab - ac - ab + b² + bc - ac + bc + c²

= a² + b² + c² - 2ab + 2bc - 2ac

Similarly, the formula of (a ± b ± c)² is in the form of a² + b² + c² ± 2ab ± 2bc ± 2ac. As you can see, a x -b = -ab, -b x -c = bc, and a x -c = -ac. Then just plug it in to the above form to get

a² + b² + c² - 2ab + 2bc - 2ac.

(a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ac

That is the faster way.

Hope this helps!