Question

Factorise Using Appropriate Identities: x^3+(y+z)^3

WITH FULL SOLUTION​

Answer 1

Step-by-step explanation:

x

3

+y

3

+z

3

−3xyz=(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)

First take L.H.S

(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)

To multiply two polynomials, we multiply each monomial of one polynomial (with its sign) by each monomial (with its sign) of the other polynomial.

x.x

2

+x.y

2

+x.z

2

−x

2

y−xyz−x

2

z+y.x

2

+y.y

2

+y.z

2

−xy

2

−y

2

z−xyz+z.x

2

+z.y

2

+z.z

2

−xyz−yz

2

−xz

2

= x

3

+xy

2

+xz

2

−x

2

y−x

2

y+yx

2

+y

3

−xy

2

−y

2

z+x

2

z+y

2

z+z

3

−yz

2

−xz

2

−3xyz

= x

3

+y

3

+z

3

−3xyz

L.H.S = R.H.S

x

3

+y

3

+z

3

−3xyz=x

3

+y

3

+z

3

−3xyz

Hence x

3

+y

3

+z

3

−3xyz=(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx) is proved.

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Answer 2

Answer:

${x}^{3} + {y}^{3} + {z}^{3} + 3 {y}^{2} z + 3y {z}^{2}$

Step-by-step explanation:

${x}^{3} + {(y + z)}^{3}$

Using identity :

${(x + y)}^{3} = {x}^{3} + {y}^{3} + 3 {x}^{2} y + 3x {y}^{2}$

We get :

${x}^{3} + ( {y}^{3} + {z}^{3} + 3 {y}^{2} z + 3y {z}^{2} )$

$= {x}^{3} + {y}^{3} + {z}^{3} + 3 {y}^{2} z + 3y {z}^{2}$