Question

expand( 5x/y + y / 5x)^3 .it's very urgent please tell me​

Answer 1

Step-by-step explanation:

We know (5x−y+z)(5x−y+z) =(5x−y+z)

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z)

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z)

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x)

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y)

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z)

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2 +2(5x)(−y)+2(−y)(z)+2(5x)(z)

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2 +2(5x)(−y)+2(−y)(z)+2(5x)(z) =25x

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2 +2(5x)(−y)+2(−y)(z)+2(5x)(z) =25x 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2 +2(5x)(−y)+2(−y)(z)+2(5x)(z) =25x 2 +y

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2 +2(5x)(−y)+2(−y)(z)+2(5x)(z) =25x 2 +y 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2 +2(5x)(−y)+2(−y)(z)+2(5x)(z) =25x 2 +y 2 +z

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2 +2(5x)(−y)+2(−y)(z)+2(5x)(z) =25x 2 +y 2 +z 2

We know (5x−y+z)(5x−y+z) =(5x−y+z) 2 Using identity (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz Here x=5x,y=−y and z=z Therefore, (5x−y+z) 2 =(5x) 2 +(−y) 2 +(z) 2 +2(5x)(−y)+2(−y)(z)+2(5x)(z) =25x 2 +y 2 +z 2 −10xy−2yz+10xz