Question

a^³+1/a^3 when a value of a =2+√3​

Answer 1

Answer:

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Answer 2

Answer:

Question

✳️

$a {}^{3} +( \frac{1}{a} ) {}^{3} when \: value \: of \: a \: = \: \\ 2 + \sqrt{3}$

Solution ⬇️

✍️ a=2+√3

✍️to find a³+1/a³

• we can give the value 'a' to find a³ + 1/a³
• first we have to find 1/a and then cube it

$= > a = 2 + \sqrt{3} \\ to \: find \: = \frac{1}{a} \\ = > \frac{1}{2 + \sqrt{3} } \\ rationalise \: denominator$

$= > \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ = > \frac{2 - \sqrt{3} }{4 - 3} ....((a - b)(a + b)) \\ or \: \frac{2 - \sqrt{3} }{1} = 2 + \sqrt{3}$

$hence \: a = 2 + \sqrt{3} \: and \: \frac{1}{a} = 2 - \sqrt{3}$

✍️ then cube it

${a}^{3} +( \frac{1}{a} ){}^{3} \\ = > (2 + \sqrt{3} ) {}^{3} + (2 - \sqrt{3} ) {}^{3} \\ = > 2 {}^{3} + ( \sqrt{3} ) {}^{3} + 3 \: \times \: 2 \: \sqrt{3} (2 + \sqrt{3} ) \\ + 2 {}^{3} -( \sqrt{3} ) {}^{3} - 3 \times 2 \times \sqrt{3}$

$((x + y){}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: and \: (x - y) {}^{3} = x {}^{3} - {y}^{3} - 3xy(x - y)) \\ formulae$

$cancel + ( \sqrt{3} ) {}^{3} \: and \: - ( \sqrt{3} ) {}^{3} \\ = > 8 + 6 \sqrt{3} (2 + \sqrt{3} ) + 8 - 6 \sqrt{3} (2 - \sqrt{3} ) \\ = > 16 + 12 \sqrt{3} + 18 - 6\sqrt{3} \\ = > 16 + 18 + 18 \\ = > 52$

✍️ Therefore a³+1/a³ = 52

Step-by-step explanation:

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