wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Factorise the following:
(x+y+z)3(2xy)3(2yz)3(2zx)3

Open in App
Solution

Let us consider 2xy+2yz+2zx=x+y+z, therefore, the given problem can be rewritten as:

[(2xy)+(2yz)+(2zx)]3(2xy)3(2yz)3(2zx)3

We know the identity: (a+b+c)3a3b3c3=3(a+b)(b+c)(c+a)

Using the above identity taking a=(2xy), b=(2yz) and c=(2zx), the equation
[(2xy)+(2yz)+(2zx)]3(2xy)3(2yz)3(2zx)3 can be factorised as follows:

(2xy)+(2yz)+(2zx)]3(2xy)3(2yz)3(2zx)3
=3((2xy)+(2yz))((2yz)+(2zx))((2xy)+(2zx))
=3(2xy+2yz)(2yz+2zx)(2xy+2zx)=3(2x+yz)(2y+zx)(xy+2z)
Hence, (x+y+z)3(2xy)3(2yz)3(2zx)3=3(2x+yz)(2y+zx)(xy+2z)


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon