Question

Without finding the cubes, factorise (x – 2y )3+ ( 2y – 3z )3 + ( 3z – x )3

Answer 1

$\huge{\mathfrak{\underline{\red{Answer!}}}}$

Hey dear!!

Question::- Find the factories of

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

Here's your answer

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*We know that,

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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

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*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)

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Hope it helps!! ☺️

Answer 2

Answer:

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Step-by-step explanation:

Hey there!

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We know,

a^3 + b^2 + 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

Here,

a = x - 2y

b = 2y - 3z

c = 3z - x

Thus,

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

And then,

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

______________

Hope it helps...!!!