Question

If 3a+b-5=0, then factorise 27a³+b³-125

Please give me the answer fast Please​

Answer 1

Answer:

   We know that 

(a+b+c)3=(a3+b3+c3)+3[(a+b+c)(ab+ac+bc)−abc]

If (a+b+c)=0, then a3+b3+c3=3abc

Given, 3a+5b+4c=0

Here, a=3a, b=5b and c=4c

Now, 

L.H.S =27a3+125b3+64c3

         =(3a)3+(5b)3+(4c)3

         =3(3a)(5b)(4c)

         =180abc

         =R.H.S

Step-by-step explanation:

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