Question

If (a+1/a)^2=9 then show that a^3+1/a^3=0

Answer 1

Step-by-step explanation:

${( \frac{a + 1}{a}) }^{2} = 9 \\ \frac{a + 1}{a} = 3 \\ 3a = a + 1 \\ 2a = 1 \\ a = \frac{1}{2}$

$to \: prove \\ \frac{ {a}^{3} + 1}{ {a}^{3} } = 0$

$\frac{a + 1}{a} = 3 \\ cubing \: both \: sides \\ {( \frac{a + 1}{a}) }^{3} = 27 \\ \frac{ {a}^{3} + {1}^{3} + 3a(a + 1) }{ {a}^{3} } = 27 \\ \frac{ {a}^{3} + 1 + 3 {a}^{2} ( \frac{a + 1}{a} ) }{ {a}^{3} } = 27 \\ \frac{ {a}^{3} + 1 + 3 {a}^{2} ( 3) }{ {a}^{3} } = 27 \\ \frac{ {a}^{3} + 1 + 9 {a}^{2} }{ {a}^{3} } = 27 \\ \frac{ {a}^{3} + 1}{ {a}^{3} } + \frac{9 {a}^{2} }{ {a}^{3} } = 27 \\ \frac{ {a}^{3} + 1}{ {a}^{3} } = 27 - \frac{9}{a}$