# Bashing unavailable - part 5

**Algebra**Level 4

Given that

\(\color{Green}{a+b+c =1} \)

\(\color{Blue}{a^2+b^2+c^2=2}\)

\(\color{Red}{a^3+b^3+c^3=3}\)

Then \(a^4+b^4+c^4= \frac{a_1}{a_2}\) .... where \(a_1\) and \(a_2\) are positive coprime integers.

And \(a^5+b^5+c^5=a_3\) ...... where \(a_3\) is an integer.

And \(a^6+b^6+c^6= \frac{a_4}{a_5}\) ....where \(a_4\) and \(a_5\) are positive coprime integers.

And \(a^7 +b^7+c^7= \frac{a_6}{a_7}\) ....where \(a_6\) and \(a_7\) are positive coprime integers.

And \(a^8+b^8+c^8 = \frac{a_8}{a_9}\) ... where \(a_8\) and \(a_9\) are positive coprime integers.

Find \(a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8 +a_9\).

**Your answer seems reasonable.**Find out if you're right!

**That seems reasonable.**Find out if you're right!

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