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