Question

Factorise [1+b3+8c3 - 6Ьc step by step verification​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:1 + {b}^{3} + {8c}^{3} - 6bc$

can be rewritten as

$\rm \:  =  \: {(1)}^{3} + {(b)}^{3} + {(2c)}^{3} - 3 \times 1 \times b \times 2c$

We know,

$\red{\sf \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz \: = } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \red{(x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)}$

So, Here,

$\red{\rm :\longmapsto\:x = 1}$

$\red{\rm :\longmapsto\:y = b}$

$\red{\rm :\longmapsto\:z = 2c}$

So, on substituting the values in above identity, we get

$\rm = (1 + b + 2c)( {1}^{2} + {b}^{2} + {(2c)}^{2} - 1 \times b - b \times 2c - 2c \times 1)$

$\rm = (1 + b + 2c)( 1 + {b}^{2} + {4c}^{2} - b - 2bc - 2c)$

More Identities to know :-

$\boxed{ \tt{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }}$

$\boxed{ \tt{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: }}$

$\boxed{ \tt{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy (x - y)\: }}$

$\boxed{ \tt{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy (x + y)\: }}$

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

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

$\boxed{ \tt{ \: {(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx) \: }}$

$\boxed{ \tt{ \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}$