Question

Simplify:

$\dfrac{1 + x + x^2+x^3+.....+x^{2018}}{1+x^3 + x^6 + ....+x^{2016}}$

Answer 1

Answer:

hello ,

Here is your solution :-

After simply the above expression we get -

{(x^2019 - 1)(x^2 + x +1)}/ (x^2017 - 1)

(x^2021 + x^2020 + x^2019-x^2-x-1)

=-------------------------------------------------------

(x^2017 - 1)

ans...

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Answer 2

Step-by-step explanation:

Since x^{2}+x+1=0x

2

+x+1=0

Consider (x^{3}-1)(x

3

−1)

Using the identity (a^{3}-b^{3})=(a-b)(a^{2}+ab+b^{2})(a

3

−b

3

)=(a−b)(a

2

+ab+b

2

)

(x^{3}-1)=(x-1)(x^{2}+x+1)(x

3

−1)=(x−1)(x

2

+x+1)

Now, \frac{(x^{3}-1)}{x-1}=(x^{2}+x+1)

x−1

(x

3

−1)

=(x

2

+x+1)

Since, (x^{2}+x+1)=0(x

2

+x+1)=0

\frac{(x^{3}-1)}{x-1}=0

x−1

(x

3

−1)

=0

{(x^{3}-1)}=0(x

3

−1)=0

{x^{3}}=1x

3

=1

so, x= 1

Now, we will find the value of x^{2015}+x^{2016}x

2015

+x

2016

= 1^{2015}+1^{2016}1

2015

+1

2016

= 1 + 1

= 2

So, the value of the given expression is 2.