Question

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

Answer 1

To Factorise :

$(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3 \\Let \: us \: take :\ x^2 - y^2 = a , y^2 - z^2 = b ;\ and :\ ^2 - x^2 = c.\\ We \: can \: observe \: an \: interesting \: property :\ here ;\\\geqslant a + b + c\\\geqslant x^2 - y^2 + y^2 - z^2 + z^2 - x^2 \\\geqslant 0\\\\\\\\So, a^3 + b^3 + c^3 = 3abc \\> 3( x^2 - y^2)(y^2-z^2)(z^2 - x^2)\\> 3(x+y)(x-y)(y+z)(y-z)(z+x)(z-x)$

This is the required answer.

____________________________________________________

Additional Information :

(a + b)² = a² + 2ab + b²  

(a + b)² = (a - b)² + 4ab  

(a - b)² = a² - 2ab + b²  

(a - b)² = (a + b)² - 4ab  

a² + b² = (a + b)² - 2ab  

a² + b² = (a - b)² + 2ab  

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²  

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)  

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)  

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

______________________________________________________________

Answer 2

$\Huge\underline{\overline{\mid{\green{Answer}}\mid}}$

Please refer to the above attachment...