Question

If a= 2+√3 then, find the value of (a6+a4+a2+1) ÷ a3

Answer 1

Answer:

answer is 56

Step-by-step explanation:

hope you find it useful

Answer 2

The value of $(a^6+a^4+a^3+a^2+1)\div a^3=56$

Step-by-step explanation:

$a=2+\sqrt 3$

$\frac{a^6+a^4+a^2+1}{a^3}$

$\frac{a^6}{a^3}+\frac{a^4}{a^3}+\frac{a^2}{a^3}+\frac{1}{a^3}$

$a^{6-3}+a^{4-3}+\frac{1}{a^{3-2}}+\frac{1}{a^3}$

Using identity:$a^x\div a^y=a^{x-y}$

$a^3+a+\frac{1}{a}+\frac{1}{a^3}$

$a^3+\frac{1}{a^3}+a+\frac{1}{a}$

$(a+\frac{1}{a})(a^2+\frac{1}{a^2}-1})+a+\frac{1}{a}$

Using identity:$a^3+b^3=(a+b)(a^2+b^2-ab)$

$(a+\frac{1}{a})(a^2+\frac{1}{a^2}-1+1)=(a+\frac{1}{a})(a^2+\frac{1}{a^2}$)

Substitute the values

$(2+\sqrt 3+\frac{1}{2+\sqrt 3})((2+\sqrt 3)^2+\frac{1}{(2+\sqrt 3)^2})$

$(\frac{(2+\sqrt 3)^2+1}{2+\sqrt 3})(4+3+4\sqrt 3+\frac{1}{4+3+4\sqrt 3})$

Using identity:$(a+b)^2=a^2+b^2+2ab$

$(\frac{4+3+4\sqrt 3+1}{2+\sqrt 3})(7+4\sqrt 3+\frac{1}{7+4\sqrt 3})$

$\frac{8+4\sqrt 3}{2+\sqrt 3}(\frac{(7+4\sqrt 3)^2+1}{7+4\sqrt 3})$

$4\times\frac{2+\sqrt 3}{2+\sqrt 3}\times \frac{49+48+56\sqrt 7+1}{7+4\sqrt 3}$

$4\times \frac{98+56\sqrt 3}{(7+4\sqrt 3)}$

$4\times 14\times (7+4\sqrt 3)\times \frac{1}{7+4\sqrt 3}$

$4\times 14=56$

The value of $(a^6+a^4+a^3+a^2+1)\div a^3=56$

#Learns more:

https://brainly.in/question/4844170:Answered  by Tomboyish