Question

Resolve into factor: {Σ_(x,y,z) x}³ - Σ_(x,y,z) x³.​

Answer 1

Explanation:

`\bf \underline{Given \:Question-} \\`

`\textsf{Resolve in to factors }`

`\sf \: {\bigg(\displaystyle\sum_(x,y,z)\rm x\bigg)}^(3) - \displaystyle\sum_(x,y,z)\rm {x}^(3)`

`\red{\large\underline{\sf{Solution-}}}`

Given expression is

`\rm :\longmapsto\: \: {\bigg(\displaystyle\sum_(x,y,z)\rm x\bigg)}^(3) - \displaystyle\sum_(x,y,z)\rm {x}^(3)`

can be rewritten as

`\rm \: = \: {(x + y + z)}^(3) - ( {x}^(3) + {y}^(3) + {z}^(3))`

can be further rewritten as

`\rm \: = \: {(x + y + z)}^(3) - {x}^(3) - {y}^(3) - {z}^(3)`

`\rm \: = \: [{(x + y + z)}^(3) - {x}^(3)] - [{y}^(3) + {z}^(3)]`

We know,

`\boxed{\tt{ {x}^(3) + {y}^(3) = (x + y)( {x}^(2) + {y}^(2) - xy)}}`

and

`\boxed{\tt{ {x}^(3) - {y}^(3) = (x - y)( {x}^(2) + {y}^(2) + xy)}}`

So, using these Identities, we get

`\rm =(x + y + z - x)[ {(x + y + z)}^(2) + {x}^(2) - x(x + y + z)] - (y + z)( {y}^(2) + {z}^(2) - yz)`

`\rm =(y + z)[ {(x + y + z)}^(2) + {x}^(2) + {x}^(2) + xy + xz] - (y + z)( {y}^(2) + {z}^(2) - yz)`

`\rm =(y + z)[ {(x + y + z)}^(2) + {2x}^(2) + xy + xz] - (y + z)( {y}^(2) + {z}^(2) - yz)`

`\rm =(y + z)[ {(x + y + z)}^(2) + {2x}^(2) + xy + xz - {y}^(2) - {z}^(2) + yz]`

`\rm =(y + z)[ {x}^(2)+{y}^(2)+{z}^(2) + 2xy + 2yz + 2zx + {2x}^(2) + xy + xz - {y}^(2) - {z}^(2) + yz]`

`\rm =(y + z)[3{x}^(2) + 3xy + 3yz + 3zx]`

`\rm \: = \: 3(y + z)( {x}^(2) + xy + yz + zx)`

`\rm \: = \: 3(y + z)[x(x + y) + z(x + y)]`

`\rm \: = \: 3(y + z)[(x + y)(x + z)]`

`\rm \: = \: 3(y + z)(x + y)(x + z)`

Hence,

`\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)}}`

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More Identities to know:-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)