# Resolve into factor: {Î£_(x,y,z) x}Â³ - Î£_(x,y,z) xÂ³.â€‹

## Answers

Step-by-step explanation:[tex] \bf \underline{Given \:Question-} \\ [/tex][tex]\textsf{Resolve in to factors }[/tex][tex] \sf \: {\bigg(\displaystyle\sum_{x,y,z}\rm x\bigg)}^{3} - \displaystyle\sum_{x,y,z}\rm {x}^{3} [/tex][tex] \red{\large\underline{\sf{Solution-}}}[/tex]Given expression is [tex]\rm :\longmapsto\: \: {\bigg(\displaystyle\sum_{x,y,z}\rm x\bigg)}^{3} - \displaystyle\sum_{x,y,z}\rm {x}^{3} [/tex]can be rewritten as [tex] \rm \: = \: {(x + y + z)}^{3} - ( {x}^{3} + {y}^{3} + {z}^{3})[/tex]can be further rewritten as [tex] \rm \: = \: {(x + y + z)}^{3} - {x}^{3} - {y}^{3} - {z}^{3}[/tex][tex] \rm \: = \: [{(x + y + z)}^{3} - {x}^{3}] - [{y}^{3} + {z}^{3}][/tex]We know, [tex]\boxed{\tt{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)}}[/tex]and [tex]\boxed{\tt{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy)}}[/tex]So, using these Identities, we get [tex] \rm =(x + y + z - x)[ {(x + y + z)}^{2} + {x}^{2} - x(x + y + z)] - (y + z)( {y}^{2} + {z}^{2} - yz)[/tex][tex] \rm =(y + z)[ {(x + y + z)}^{2} + {x}^{2} + {x}^{2} + xy + xz] - (y + z)( {y}^{2} + {z}^{2} - yz)[/tex][tex] \rm =(y + z)[ {(x + y + z)}^{2} + {2x}^{2} + xy + xz] - (y + z)( {y}^{2} + {z}^{2} - yz)[/tex][tex] \rm =(y + z)[ {(x + y + z)}^{2} + {2x}^{2} + xy + xz - {y}^{2} - {z}^{2} + yz][/tex][tex] \rm =(y + z)[ {x}^{2}+{y}^{2}+{z}^{2} + 2xy + 2yz + 2zx + {2x}^{2} + xy + xz - {y}^{2} - {z}^{2} + yz][/tex][tex] \rm =(y + z)[3{x}^{2} + 3xy + 3yz + 3zx][/tex][tex] \rm \: = \: 3(y + z)( {x}^{2} + xy + yz + zx)[/tex][tex] \rm \: = \: 3(y + z)[x(x + y) + z(x + y)][/tex][tex] \rm \: = \: 3(y + z)[(x + y)(x + z)][/tex][tex] \rm \: = \: 3(y + z)(x + y)(x + z)[/tex]Hence, [tex] \boxed{\tt{ \sf {\bigg(\displaystyle\sum_{x,y,z}\rm x\bigg)}^{3} - \displaystyle\sum_{x,y,z}\rm {x}^{3} = 3(x + y)(y + z)(z + x)}}[/tex]â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬â–¬More Identities to know:-(a + b)Â² = aÂ² + 2ab + bÂ²(a - b)Â² = aÂ² - 2ab + bÂ²aÂ² - bÂ² = (a + b)(a - b)(a + b)Â² = (a - b)Â² + 4ab(a - b)Â² = (a + b)Â² - 4ab(a + b)Â² + (a - b)Â² = 2(aÂ² + bÂ²)(a + b)Â³ = aÂ³ + bÂ³ + 3ab(a + b)(a - b)Â³ = aÂ³ - bÂ³ - 3ab(a - b)