Question

a+1/a =1 then a^3+1=?​

Answer 1

Given :-

$\boxed{ \huge \rm{}a + \frac{1}{a} = 1}$

To find :- $\bf {a}^{3 + 1}$

SOLUTION :-

$\bf \huge{a + \frac{1}{a} = 1} \\ \\ \rm \underline{ \overline{ cubing \: \: both \: \: sides}} \\ \\ = > \tt {a}^{3} + \frac{1}{ {a}^{3} } + 3( {\cancel{a}}) \bigg( \frac{1}{ \cancel{a}} \bigg) \bigg[a + \frac{1}{a} \bigg] = {1}^{3} \\ \\ \tt= > {a}^{3} + \frac{1}{ {a}^{3} } + 3\bigg[a + \frac{1}{a} \bigg] = 1 \\ \\ \tt = > \: \: \: \: \: \: \: {a}^{3} + \frac{1}{ {a}^{3} } + 3(1) = 1 \\ \\ = > \tt \: \: \: \: \: \: \: \: \: \: \: \: \: {a}^{3} + \frac{1}{ {a}^{3} } = 1 - 3 \\ \\ = > \boxed{\boxed{\tt \: \: \: \: \: \: \: \: \: \: \: {a}^{3} + \frac{1}{ {a}^{3} } = - 2 \: \: \: \: }}$

Also,

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