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

Add following.
(i) abbc,bcca,caab
(ii) ab+ab,bc+bc,ca+ac
(iii) 2p2q23pq+4,5+7pq3p2q2
(iv) l2+m2,m2+n2,n2+l2,2lm+2mn+2nl

Open in App
Solution

(i) Given (abbc)+(bcca)+(caab)
=abbc+bcca+caab
Grouping the like terms as
=abba+bcbc+caca
All terms gets cancelled, So
=0
Hence, the addition of given expression is 0


(ii) Given (ab+ab)+(bc+bc)+(ca+ac)
=ab+ab+bc+bc+ca+ac
Grouping the like terms as
=aa+bb+cc+ab+bc+ca
By simplifying, we get
=ab+bc+ac

(iii) Given (2p2q23pq+4)+(5+7pq3p2q2)
=2p2q23pq+4+5+7pq3p2q2
Grouping the like terms, we get
=2p2q23p2q2+7pq3pq+5+4
By simplifying, we get
=p2q2+4pq+9

(iv) Given (l2+m2)+(m2+n2)+(n2+l2)+(2lm+2mn+2nl)
Grouping the like terms, we get
=l2+l2+m2+m2+n2+n2+2lm+2mn+2nl
=2l2+2m2+2n2+2(lm+mn+nl)
By simplifying, we get
=2(l2+m2+n2+lm+mn+nl)

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Method of Common Factors
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon