Find the value of a6-1-a3 given that a2-1-a-2

The correct option is B 14(a^2 1)^3 = a^6 3a^4 + 3a^2 1 =(a^6 1) + 3a^2(1 a^2) a^6 1/a^3 + 3(1 a^2)/a = 8 a^6 1/a^3 = 14 ...

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Question

Find the value of $\frac{a^{6} - 1}{a^{3}}$ given that $\frac{a^{2} - 1}{a} = 2$

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\frac{a^{6} - 1}{a^{3}}

\frac{a^{2} - 1}{a} = 2

\frac{a^{2} - 1}{a} = 2

\frac{a^{6} - 1}{a^{3}}

a_{1}, a_{2}, a_{3}, \dots

a_{1} + a_{2} + a_{3} + \dots \dots + a_{98} = 137

a_{2} + a_{4} + a_{6} + \dots \dots + a_{98}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} a n d a_{6}

a_{6} > a_{1}, a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = p^{3} a n d a_{5} = 39

(a_{3} + a_{4} - 12 p)

a_{1}, a_{2}, a_{3}, . . . ., a_{n}, . . .

a_{n} = \frac{a_{n - 1} + a_{n - 2}}{2}

n \geq 3

a_{3} = 4

a_{5} = 20

a_{6}

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