Math, asked by a1aysridh3uri, 1 year ago

Factorize (x-3y)^3+(3y-7z)^3+(7z-x)^3

Answers

Answered by mindfulmaisel
60

"(x-3y)^{ 3 }\quad +\quad (3y-7z)^{ 3 }\quad +\quad (7z-x)^{ 3 }\quad =\quad 3(x-3y)\quad (3y-7z)\quad (7z-x)

Given:

(x-3y)^{ 3 }\quad +\quad (3y-7z)^{ 3 }\quad +\quad (7z-x)^{ 3 }

To find:

Factorize (x-3y)^{ 3 }\quad +\quad (3y-7z)^{ 3 }\quad +\quad (7z-x)^{ 3 }

Solution:

“We have to factorize”,

(x-3y)^{ 3 }\quad +\quad (3y-7z)^{ 3 }\quad +\quad (7z-x)^{ 3 }

Let a = x-3y ; b = 3y-7z ; c = 7z-x

now,

a + b + c=x-3y + 3y-7z + 7z-x = 0

Since, a + b + c = 0,

then we have,

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

Hence,

(x-3y)^{ 3 }\quad +\quad (3y-7z)^{ 3 }\quad +\quad (7z-x)^{ 3 }\quad =\quad 3(x-3y)\quad (3y-7z)\quad (7z-x)"

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