Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find (x + 1)^6 + (x - 1)^6 . Hence or otherwise evaluate (root of 2 + 1)^6 + ( root of 2 - 1) ^ 6 .

Q12. Find (x+1)^6 + (x-1)^6. Hence or otherwise evaluate (\sqrt2+1)^6 + (\sqrt2-1)^6.

Answers (1)

Using the binomial theorem, the expressions (x+1)^4 and (x-1)^4 can be expressed as

(x+1)^6=^6C_0x^6+^6C_1x^51+^6C_2x^41^2+^4C_3x^31^3+^6C_4x^21^4+^6C_5x1^5+^6C_61^6

(x-1)^6=^6C_0x^6-^6C_1x^51+^6C_2x^41^2-^4C_3x^31^3+^6C_4x^21^4-^6C_5x1^5+^6C_61^6

From here,

\\(x+1)^6-(x-1)^6=^6C_0x^6+^6C_1x^51+^6C_2x^41^2+^4C_3x^31^3+^6C_4x^21^4+^6C_5x1^5+^6C_61^6\:\:\:\:\;\:\:\;\:\:\:\ +^6C_0x^6-^6C_1x^51+^6C_2x^41^2-^4C_3x^31^3+^6C_4x^21^4-^6C_5x1^5+^6C_61^6

(x+1)^6+(x-1)^6=2(^6C_0x^6+^6C_2x^41^2+^6C_4x^21^4+^6C_61^6)

(x+1)^6+(x-1)^6=2(x^6+15x^4+15x^2+1)

Now, using this, we get 

(\sqrt2+1)^6 + (\sqrt2-1)^6=2((\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1)

(\sqrt2+1)^6 + (\sqrt2-1)^6=2(8+60+30+1)=2(99)=198

Posted by

neha

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE 12th Result 2026 Date: Here’s what previous years’ trends indicate
anam-khan

CBSE Class 12th marks uploading for schools in West Asia starts from April 8
anam-khan

CBSE question papers QR codes part of 'internal systems'; board cautions 'unverified claims'

Latest Question