Question

If a²+1/a²=47 and a is not equal to 0 then find a³+1/a³

Answer 1

$\bf \red Solutio \red {n}$

$(a + \frac{1}{a} ) {}^{2} = {a}^{2} + \frac{1}{a {}^{2} } + 2 \times a \times \frac{1}{a}$

$\rightarrow \: (a + \frac{1}{a} ) {}^{2} = a {}^{2} + \frac{1}{a {}^{2} } + 2$

$\rightarrow \: (a + \frac{1}{a} ) {}^{2} = 47 + 2$

$\rightarrow \: (a + \frac{1}{a} ) {}^{2} = 49$

$\rightarrow(a + \frac{1}{a} ) = \sqrt{49}$

$\rightarrow \red {a + \frac{1}{a} = 7}$

$\star \: \orange{Cubing \: both \: sides }$

$(a + \frac{1}{a} ) {}^{3} =( 7) {}^{3}$

$\rightarrow{a}^{3} + \frac{1}{ {a}^{3} } + 3(a + \frac{1}{a} ) = 343$

$\rightarrow \: {a}^{3} + \frac{1}{a {}^{3} } + 3 \times 7 = 343$

$\rightarrow \: {a}^{3} + \frac{1}{ {a}^{3} } = 343 - 21$

$\rightarrow \blue{{a}^{3} + \frac{1}{ {a}^{3} } }= \purple{ 322}$

Answer 2

Given :

${a}^{2} + \dfrac{1}{ {a}^{2} } = 47$

a ≠ 0

To find :

${a}^{3} + \dfrac{1}{ {a}^{3} }$

Solution :

By using identity : (a + b)² = a² + b² + 2ab

=> ${(a + \dfrac{1}{a}) }^{2} = {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times a \times \dfrac{1}{a}$

=> ${(a + \dfrac{1}{a}) }^{2} = {a}^{2} + \dfrac{1}{ {a}^{2} } + 2$

=> ${(a + \dfrac{1}{a}) }^{2}$ = 47 + 2

=> ${(a + \dfrac{1}{a}) }^{2} = 49$

=> $(a + \dfrac{1}{a}) = \sqrt{49}$

=> $(a + \dfrac{1}{a}) = 7$

Now, cubing both sides

=> ${(a + \dfrac{1}{a}) }^{3} = {(7)}^{3}$

Using identity : (a + b)³ = a³ + b³ + 3ab(a + b)

=> ${a}^{3} + \dfrac{1}{ {a}^{3} } + 3 \times a \times \dfrac{1}{a}(a + \dfrac{1}{a}) = 343$

=> ${a}^{3} + \dfrac{1}{ {a}^{3} } + 3 \times 7 = 343$

=> ${a}^{3} + \dfrac{1}{ {a}^{3} } = 343 - 21$

=> ${a}^{3} + \dfrac{1}{ {a}^{3} } = 322$

Answer :

${a}^{3} + \dfrac{1}{ {a}^{3} } = 322$