Question

factorise the polynomial a square minus b square whole cube plus b square minus c square whole cube plus c square minus a square whole cube​

Answer 1

Explanation:

From what you have written, this is the polynomial that makes most sense to me:

$\mathsf{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}$

$\mathsf{=x^3+y^3+z^3}$ [substituting the terms with x, y and z]

$\mathsf{=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz}$ [according to formula]

$\mathsf{=(a^2-b^2+b^2-c^2+c^2-a^2)(x^2+y^2+z^2-xy-yz-zx)+3xyz}$ [substituting the values of x, y and z in the first term]

$\mathsf{=0(x^2+y^2+z^2-xy-yz-zx)+3xyz}$

$\mathsf{=0+3xyz=3xyz}$

$\mathsf{=3(a^2-b^2)(b^2-c^2)(c^2-a^2)}$

Answer: 3(a^2-b^2)(b^2-c^2)(c^2-a^2)

I hope this helps. :D

Answer 2

Answer:

From what you have written, this is the polynomial that makes most sense to me:

(

�

2

−

�

2

)

3

+

(

�

2

−

�

2

)

3

+

(

�

2

−

�

2

)

3

(a

2

−b

2

)

3

+(b

2

−c

2

)

3

+(c

2

−a

2

)

3

=

�

3

+

�

3

+

�

3

=x

3

+y

3

+z

3

[substituting the terms with x, y and z]

=

(

�

+

�

+

�

)

(

�

2

+

�

2

+

�

2

−

�

�

−

�

�

−

�

�

)

+

3

�

�

�

=(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)+3xyz [according to formula]

=

(

�

2

−

�

2

+

�

2

−

�

2

+

�

2

−

�

2

)

(

�

2

+

�

2

+

�

2

−

�

�

−

�

�

−

�

�

)

+

3

�

�

�

=(a

2

−b

2

+b

2

−c

2

+c

2

−a

2

)(x

2

+y

2

+z

2

−xy−yz−zx)+3xyz [substituting the values of x, y and z in the first term]

=

0

(

�

2

+

�

2

+

�

2

−

�

�

−

�

�

−

�

�

)

+

3

�

�

�

=0(x

2

+y

2

+z

2

−xy−yz−zx)+3xyz

=

0

+

3

�

�

�

=

3

�

�

�

=0+3xyz=3xyz

=

3

(

�

2

−

�

2

)

(

�

2

−

�

2

)

(

�

2

−

�

2

)

=3(a

2

−b

2

)(b

2

−c

2

)(c

2

−a

2

)

Answer: 3(a^2-b^2)(b^2-c^2)(c^2-a^2)

I hope this helps. :D