Question

Factorisex3+3x2y+3xy2+y3–125​

Answer 1

Answer:

x3+3x2y+3xy2+y3–125​=0

⇒(x3+3x2y+3xy2+y3)=125

⇒[x3+3xy(x+y)+y3]=125  

Using the identity, a

3 +3ab(a+b)+b

3 =(a+b)

3

we have

x 3 +3xy(x+y)+y 3 =(x+y)

3

⇒(x+y) 3 =5 3

⇒(x+y) 3 −5 3

is of the form a

3 −b 3

=(a−b)(a 2 +ab+b 2)

 

=(x+y−5)((x+y) 2 +5(x+y)+5 2

=(x+y−5)(x  +y 2 +2xy+5x+5y+25)

Step-by-step explanation:

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