The expression a3 - 1a3 equals

The correct option is D ( a 1a ) ( a^2 + 1 + 1a^2 ) a ^ 3 1 a ^ 3 =( a 1 a #160; ) ( a ^ 2 +a· 1 a + 1 a ^ 2 #160; ) =( a 1 a ...

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Expansion of (a ± b ± c)²

The expressio...

Question

The expression $a^{3} - {\frac{1}{a}}^{3}$ equals

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

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

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

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

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

a_{1}, a_{2}, a_{3} . . . . a_{15} \in R^{+}

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

(1 + a_{1} + a_{1}^{2}) (1 + a_{3} + a_{3}^{2}) . . . . (1 + a_{a} + a_{a}^{2})

a = 6 - \sqrt{35},

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

(b) a^{2} - \frac{1}{a^{2}}

(a) a^{4} - \frac{1}{a^{4}}

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

a_{1}, a_{2}, a_{3}, a_{4}

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}

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