Question

Factorize each of the following expressions:
27x³-y³-z³-9xyz

Answer 1

The algebraic expression is given as: 27x³ - y³ - z³ - 9xyz

 

We can rewrite the  given expression as :  

= (3x)³ + (- y)³ + (- z)³ − 3 × 3x × (- y)(- z)

In order to factorise the given algebraic expression of the form a³ + b³ + c³ - 3abc , we use the following Identity -: a³ + b³ + c³ - 3abc = (a + b + c) (a² + b³ + c² - ab - bc - ca) :  

= {(3x - y - z)}{(3x)² + (-y)² + (-z)² − 3x(- y) − (- y)(-z) −(-z)(3x)}

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

Hence, the factorization of an algebraic expression is (3x - y - z)(9x² + y²  + z² + 3xy - yz + 3zx).

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Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

$\rule{200}{2}$

$\sf{\ddag \: \: \: \: \: \: Equation : \: \: 27x^3 - y^3 - z^3 - 9xyz \: \: \: \: \: \: }$

As, we have to factorise the Equation.

We can re - write it.

$\sf{→(3x)^3 + (-y)^3 + (-z)^3 - 3(3x)(-y)(-z)} \\ \\ \bf{We \: know \: that,} \\ \\ \footnotesize{\star{\boxed{\sf{a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca}}}}$

Where,

a is 3x

b is -y

c is -z

$\rule{150}{2}$

$\sf{→(3x - y - z)\bigg((3x)^2 + (-y)^2 + (-z)^2 - (3x)(-y) - (-y)(-z) - (3x)(-z) \bigg)} \\ \\ \sf{→(3x - y - z)(9x^2 + y^2 + z^2 + 3xy - yz + 3xz)}$