Question

Factorise :


-Brainly​

Answer 1

Answer:          $= (3x+2y+1z)[9x^2+4y^2+z^2-6xy-2yz-3zx]$

                               

Step-by-step explanation:

Since we know that

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

so, rewriting  THIS:

$27x^3 + 8y^3 + z^3 - 3abc$

In this type:

$a^3 + b^3 + c^3 - 3abc$

As follows:

$3x^3 + 2y^3 + 1z^3 - 3(3x)(2y) (1z)$

Now, using this identity:

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

$=3x^3 + 2y^3 + 1z^3 - 3(3x)(2y) (1z)\\= (3x+2y+1z)[(3x)^2+(2y)^2+(z)^2-(3x)(2y)-(2y)(1z)-(1z)(3x)]$

$= (3x+2y+1z)[9x^2+4y^2+z^2-6xy-2yz-3zx]$

$Learning\;\; together\;:)$

Answer 2

Answer:

(3x+2y+z) (9x2+4y2+z2-6xy-2yz-3zx)

Step-by-step explanation:

27x3+8y3+z3-18xyz

it is in the form

a3+b3+c3-3abc= (a+b+c) (a2+b2+c2-ab-bc-ca)

we can write it as

(3x)3+(2y)3+z3-3*2*3* xyz

=>(3x+2y+z) (9x2+4y2+z2-6xy-2yz-3zx)