Question

If a (all over) a^2 + 1 = 1/3, determine a^3 (all over) a^6+a^5+a^4+a^3+a^2+1

Will be giving medals to the one who can help me~~~

Answer 1

\[If \frac{ a }{ a^{2} + 1 } = \frac{ 1 }{ 3 }, determine \frac{ a^{3} }{ a^{6} + a^{5} + a ^{4} + a ^{3} + a ^{2} +1 }\]

Answer 2

a/a^2+1 = 1/3
3a/a^2+1=1

a=a^2+1/3

substitute "a" in the 2nd equation

Answer 3

Hmm. Won't that take a lot of time? </3

Answer 4

\[\frac{a}{a^2+1}=\frac{1}{3}\\\frac{a^2+1}{a}=3\\a+\frac{1}{a}=3\] it maybe help

Answer 5

\[\frac{a^3}{a^6+a^5+a^4+a^3+a^2+1}=\\\frac{a^3}{(a^6+a^4)+(a^5+a^3)+(a^2+1)\\\frac{a^3}{(a^2+1)(a^4+a^3+1)}}=\]

Answer 6

That's a big help amoodarya! I hope I can use it to solve this problem.

Answer 7

\[\frac{a^3}{(a^2+1)(a^4+a^3+1)}=\\\]

Answer 8

you go from here !
but I feel denominator must have "a"
i.e. a^6+a^5+...+a^2+a+1

Answer 9

Well, thanks for your great help! Never realized it was factorable. Hihi

Answer 10

Sooo, I got:

\[\frac{ a }{ a ^{2} + 1} \times \frac{ a ^{2} }{ a ^{4} + a ^{3} + 1 }\]

which is equal to

\[\frac{ 1 }{ 3 } \times \frac{ a ^{2} }{ a ^{4} + a ^{3} + 1 }\]

Answer 11

I'll just check this out again later. Haha.

Answer 12

\[\frac{1}{3}\frac{1}{\frac{a^4+a^3+1}{a^2}}\]

Answer 13

Hmm. Thanks. The answer was \[\frac{ 1 }{ 29 }\] but I really had no idea why. Haha