Question

factorise the following plzzz​

Answer 1

Answer:

sorry dont have answer

Step-by-step explanation:

Answer 2

Answer:

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

Step-by-step explanation:

In the question,

We have to factorise the term,

$(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}$

Now, we know that the properties of the sum of the cubes,

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

So,

We can see here that,

a = (x - 2y)

b = (2y - 3z)

c = (3z - x)

So,

a + b + c = (x - 2y) + (2y - 3z) + (3z - x) = 0

So,

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\\a^{3}+b^{3}+c^{3}-3abc=(0)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\\a^{3}+b^{3}+c^{3}-3abc=0\\a^{3}+b^{3}+c^{3}=3abc$

So,

We can say that,

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

Therefore, the factorised equation is given by,

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