Question

Without finding the cubes facterise.
x-2y)² + (2y-32 3 + (32)3
​

Answer 1

Answer:

Step-by-step explanation:

- Find the factories of

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

Here's your answer

==================>>>

*We know that,

______________________

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

If,

$(a + b + c) = 0$

Then,

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

______________________

*And here,

If,

$a = (x - 2y)$

$b = (2y - 3z)$

$c = (3z - x)$

Thus,

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

And last,

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

_________________

Hope it helps!!

:)

Answer 2

Given to factorise:-

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

Answer:-

We know that,

If,

$a + b + c = 0$

Then,

${a}^{3} + {b}^{3} + {c}^{3} = 3abc$

Now,

$(x - 2y) + (2y - 3z) + (3z - x) = 0$

Therefore,

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

$<hr/>$

Answer:-

$3(x - 2y)(2y - 3z)(3z - x)$

$<marquee>Hope it helps you...Please mark this answer as the brainliest and follow me..</marquee>$