Question

factorise xcube-ycube

Answer 1

Answer:

$\bf\x^3-y^3=(x-y)(x^2+xy+y^2)$

Step-by-step explanation:

Given:

$x^3-y^3$

We know that

$\boxed{\bf(a-b)^3=a^3-b^3-3ab(a-b)}$

$(x-y)^3=x^3-y^3-3xy(x-y)$

$x^3-y^3=(x-y)^3+3xy(x-y)$

$x^3-y^3=(x-y)[(x-y)^2+3xy]$

$x^3-y^3=(x-y)[x^2-2xy+y^2+3xy]$

$\implies\boxed{\bf\;x^3-y^3=(x-y)(x^2+xy+y^2)}$

Answer 2

Answer: .

Step-by-step explanation:

x

3

−y

3

=(x−y)(x

2

+xy+y

2

)

Step-by-step explanation:

Given:

x^3-y^3x

3

−y

3

We know that

\boxed{\bf(a-b)^3=a^3-b^3-3ab(a-b)}

(a−b)

3

=a

3

−b

3

−3ab(a−b)

(x-y)^3=x^3-y^3-3xy(x-y)(x−y)

3

=x

3

−y

3

−3xy(x−y)

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

3

−y

3

=(x−y)

3

+3xy(x−y)

x^3-y^3=(x-y)[(x-y)^2+3xy]x

3

−y

3

=(x−y)[(x−y)

2

+3xy]

x^3-y^3=(x-y)[x^2-2xy+y^2+3xy]x

3

−y

3

=(x−y)[x

2

−2xy+y

2

+3xy]

\implies\boxed{\bf\;x^3-y^3=(x-y)(x^2+xy+y^2)}⟹

x

3

−y

3

=(x−y)(x

2

+xy+y

2

)