Computer Science, asked by anujaak123, 3 months ago

Factorise each of the following i. 27-125a³-135a+225a² ii. 27p³-½16-9/2p²+¼p

Answers

Answered by Anonymous
11

Explanation:

27 – 125a³– 135a + 225a²

27 – 125a³– 135a + 225a²= (3)³ – (5a)³ – 135a + 225a²

27 – 125a³– 135a + 225a²= (3)³ – (5a)³ – 135a + 225a²= (3)³ – (5a)³ – 45a(3 – 5a)

27 – 125a³– 135a + 225a²= (3)³ – (5a)³ – 135a + 225a²= (3)³ – (5a)³ – 45a(3 – 5a)= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)

27 – 125a³– 135a + 225a²= (3)³ – (5a)³ – 135a + 225a²= (3)³ – (5a)³ – 45a(3 – 5a)= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)Again Using (x – y)³ = x³ – y³ – 3xy (x – y)

27 – 125a³– 135a + 225a²= (3)³ – (5a)³ – 135a + 225a²= (3)³ – (5a)³ – 45a(3 – 5a)= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)Again Using (x – y)³ = x³ – y³ – 3xy (x – y)Putting nd we get x = 3, y = 5a

27 – 125a³– 135a + 225a²= (3)³ – (5a)³ – 135a + 225a²= (3)³ – (5a)³ – 45a(3 – 5a)= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)Again Using (x – y)³ = x³ – y³ – 3xy (x – y)Putting nd we get x = 3, y = 5a= (3 – 5a)³

27 – 125a³– 135a + 225a²= (3)³ – (5a)³ – 135a + 225a²= (3)³ – (5a)³ – 45a(3 – 5a)= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)Again Using (x – y)³ = x³ – y³ – 3xy (x – y)Putting nd we get x = 3, y = 5a= (3 – 5a)³= (3 – 5a) (3 – 5a) (3 – 5a)

I hope it is helpful for you ❤️

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