Question

x³-3x²y+3xy²-2y³.Factorise​

Answer 1

Step-by-step explanation:

    Now, x³ + 3x²y + 3xy² + 2y³

    = ( x³ + 3x²y + 3xy² + y³ ) + y³

    = ( x + y )³ + y³

    = ( x + y + y ) { ( x + y )² - ( x + y ) y + y² }

    = ( x + 2y ) ( x² + 2xy + y² - xy - y² + y² )

    = ( x + 2y ) ( x² + xy + y² )

which is the required factorization

Algebraic Identities :

  ( a + b )³ = a³ + 3a²b + 3ab² + b³

  a³ + b³ = ( a + b ) ( a² - ab + b² )

hope it helps

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Answer 2

Answer:

(x+y)³ = x³ + 3x²y + 3xy² + y³(x-y)³ = x³ - 3x²y + 3xy² - y³(y-x)³ = y³ - 3y²x + 3yx² -x³ (x+1)³ = x³ + 3x² + 3x + 1(x-1)³ = x³ - 3x² + 3x -1(y+1)³ = y³ +3y² + 3y +1(y-1)³ = y³ - 3y² + 3y -1

thanks you ..