Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
903 views
in Algebraic Expressions by (56.3k points)

Simplify:

(i) (a + b + c)2 + (a − b + c)2 

(ii) (a + b + c)2 − (a − b + c)2 

(iii) (a + b + c)2 + (a – b + c)2 + (a + b − c)2 

(iv) (2x + p − c)2 − (2x − p + c)2 

(v) (x2 + y2 − z2)2 − (x2 − y2 + z2)2

1 Answer

0 votes
by (30.5k points)
selected by
 
Best answer

(i) (a + b + c)2 + (a − b + c)2 

= (a2 + b2 + c2 + 2ab+2bc+2ca) + (a2 + (−b)2 + c2 −2ab−2bc+2ca) 

= 2a2 + 2b2 + 2c2 + 4ca 

(ii) (a + b + c)2 − (a − b + c)2 

= (a2 + b2 + c2 + 2ab + 2bc + 2ca) − (a2 + (−b)2 + c2 − 2ab − 2bc + 2ca) 

= a2 + b2 + c2 + 2ab + 2bc + 2ca − a2 − b2 − c2 + 2ab + 2bc − 2ca 

= 4ab + 4bc 

(iii) (a + b + c)2 + (a – b + c)2 + (a + b − c)2 

= a2 + b2 + c2 + 2ab + 2bc + 2ca + (a2 + b2 + (c)2 − 2bc − 2cb + 2ca) + (a2 + b2 + c2 + 2ab − 2bc – 2ab) 

= 3a2 + 3b2 + 3c2 + 2ab − 2bc + 2ca 

(iv) (2x + p − c)2 − (2x − p + c)2 

= [4x2 + p2 + c2 + 4xp − 2pc − 4xc] − [4x2 + p2 + c2 − 4xp− 2pc + 4xc] 

= 4x2 + p2 + c2 + 4xp − 2pc − 4cx − 4x2 − p2 − c2 + 4xp + 2pc− 4cx

= 8xp − 8xc 

= 8(xp − xc)

(v) (x2 + y2 − z2)2 − (x2 − y2 + z2)2 

= (x2 + y2 + (−z)2)2 − (x2 − y2 + z2)2 

= [x4 + y4 + z4 + 2x2y2 + 2y2z2 + 2x2z2 − [x4 + y4 + z4 − 2x2y2 − 2y2z2 + 2x2z2

= 4x2y2 – 4z2x2

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...