1a(a2+1)=3,1a(a2+1)=3,\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}    then the value of a6+1a3" role="presentation">a6+1a3a6+1a3\frac{{{a^6} + 1}}{{{a^3}}}   = ?" /> 1a(a2+1)=3,1a(a2+1)=3,\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}    then the value of a6+1a3" role="presentation">a6+1a3a6+1a3\frac{{{a^6} + 1}}{{{a^3}}}   = ?" /> 1a(a2+1)=3,1a(a2+1)=3,\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}    then the value of a6+1a3" role="presentation">a6+1a3a6+1a3\frac{{{a^6} + 1}}{{{a^3}}}   = ?" />

If
1a(a2+1)=3,
   then the value of
a6+1a3
  = ?

  • 19
  • 218
  • 327
  • 41
Answer:- 1
Explanation:-

Solution:
1a(a2+1)=3a+1a=3a3+1a3+3.a.1a(a+1a)=33a3+1a3+3(3)=33a3+1a3=279a3+1a3=18a6+1a3=18

Post your Comments

Your comments will be displayed only after manual approval.

   then the value of
a6+1a3
  = ?", "text": "If
1a(a2+1)=3,
   then the value of
a6+1a3
  = ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "
Solution:
1a(a2+1)=3a+1a=3a3+1a3+3.a.1a(a+1a)=33a3+1a3+3(3)=33a3+1a3=279a3+1a3=18a6+1a3=18
", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "
Solution:
1a(a2+1)=3a+1a=3a3+1a3+3.a.1a(a+1a)=33a3+1a3+3(3)=33a3+1a3=279a3+1a3=18a6+1a3=18
", "dateCreated": "7/24/2019 10:09:12 AM" } }
Test
Classes
E-Book