Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
573 views
in Binomial Theorem by (48.0k points)

Expand (a + b)4 – (a – b)4. Using this find the value of (√3 + √2)4 – (√3 – √2)4

1 Answer

+1 vote
by (46.3k points)
selected by
 
Best answer

By using binomial theorem,

(a + b)4 = 4C0 ab0 + 4C1 ab1
4C2 ab24C3 ab3 + 4Cab4

4C0 a4 + 4C1 ab1 + 4C2 ab2
4C3 ab3 + 4Cb4 …. (1)

and (a – b)4 = 4C0 a4 (-b)04C1 a3 (-b)1
4C2 a2 (-b)2 + 4Ca1 (-b)3 + 4Ca0 (-b)4

4C0 a4 – 4C1 a3 b + 4C2 a2 b2
– 4C3 ab4 + 4C4 b4 …. (2)

From equation (1) and (2) we have,

(a + b)4 – (a – b)4

= [4C0 a4 + 4C1 ab +4C2 a2 b2 + 4C3 ab3 + 4C4 b4]
– [4C0 a4 – 4Cab + 4Cab2 –4C3 ab3 + 4Cb4]

4Ca4 + 4C1 ab + 4Cab2 + 4Cab3 + 4Cb4
– 4Ca4 + 4Ca3b – 4C2 a2b2 + 4Cab3 – 4Cb4

= 2. 4C1 a3 b + 2. 4C3 ab3

= 2ab [4C1 a2 + 4Cb2]

= 2ab [4a2 + 4b2]  [∴ 4C1 = 4, 4C3 = 4]

= 2ab (a2 + b2)

Hence, (a + b)4 – (a – b)4 = 8ab (a2 + b2)

Now, putting a= √3 and b = √2

(√3 + √2)4 – (√3 – √2)4

= 8 √3 × √2 [(√3)2 + (√2)2]

= 8 √6 (3 + 2)

= 8 √6 × 5 – 40 √6

Hence (√3 + √2 )4 – (√3 – √2 )4 = 40 √6

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.

...