Question

Resolve The factors :
Please don't Spam
Answer with full explanation
​

Answer 1

Answer:

x=x y z+x=3x

ans is 3x

Step-by-step explanation:

please mark me brainliest

Answer 2

Given Question

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,

$\boxed{\tt{ \sf {\bigg(\displaystyle\sum_{x,y,z}\rm x\bigg)}^{3} - \displaystyle\sum_{x,y,z}\rm {x}^{3} = 3(x + y)(y + z)(z + x)}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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)