While I can see a process involving determining the values of “a”, “a²”, “a³” and even “a⁴” to several decimal places, that process is likely to be time-consuming and inherently error-prone due to probable rounding.
My approach as shown below is better in every respect.
Consider ( a + 1/a )²
= ( a + 1/a )( a + 1/a ) =
= ( a² + 1 + 1 + 1/a² ) =
= ( a² + 1/a² ) + 2 Equation 1
Consider ( a + 1/a )³ =
= ( a + 1/a )( 1 + 1/a )( a + 1/a ) or from Equation 1 =
= ( a + 1/a )( a² + 1/a² + 2 ) =
= ( a³ + 1/a +2a + a +1/a³ + 2/a ) =
= ( a³ + 1/a³ ) + ( 3a + 3/a ) =
= ( a³ + 1/a³ ) + 3( a + 1/a ) = =Equation 2
Consider ( a² + 1/a² )²
= ( a² + 1/a² )( a² + 1/a² ) =
= ( a⁴ + 1/a⁴ + 1 + 1 ) =
= ( a⁴ + 1/a⁴ ) + 2 = 47 (given) or
= ( a⁴ + 1/a⁴ ) = 47 + 2 = 49
Therefore ( a² + 1/a² )² = 49
If we take the square root of both sides and note by definition that the left hand side must be positive due to a² and 1/a², then:
√( a² + 1/a² ) = √(49) or (a²+1/a²) = +7 (ignoring the negative output of the square root).
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Answer:
While I can see a process involving determining the values of “a”, “a²”, “a³” and even “a⁴” to several decimal places, that process is likely to be time-consuming and inherently error-prone due to probable rounding.
My approach as shown below is better in every respect.
Consider ( a + 1/a )²
= ( a + 1/a )( a + 1/a ) =
= ( a² + 1 + 1 + 1/a² ) =
= ( a² + 1/a² ) + 2 Equation 1
Consider ( a + 1/a )³ =
= ( a + 1/a )( 1 + 1/a )( a + 1/a ) or from Equation 1 =
= ( a + 1/a )( a² + 1/a² + 2 ) =
= ( a³ + 1/a +2a + a +1/a³ + 2/a ) =
= ( a³ + 1/a³ ) + ( 3a + 3/a ) =
= ( a³ + 1/a³ ) + 3( a + 1/a ) = =Equation 2
Consider ( a² + 1/a² )²
= ( a² + 1/a² )( a² + 1/a² ) =
= ( a⁴ + 1/a⁴ + 1 + 1 ) =
= ( a⁴ + 1/a⁴ ) + 2 = 47 (given) or
= ( a⁴ + 1/a⁴ ) = 47 + 2 = 49
Therefore ( a² + 1/a² )² = 49
If we take the square root of both sides and note by definition that the left hand side must be positive due to a² and 1/a², then:
√( a² + 1/a² ) = √(49) or (a²+1/a²) = +7 (ignoring the negative output of the square root).
Therefore (a² + 1/a²) = 7Equation 3
Consider Equations 1 and 2
(a + 1/a)(a + 1/a) = (a² + 1/a²) + 2 or
(a + 1/a)² = (a² + 1/a²) + 2 or from Equation 3
(a + 1/a)² = ( 7 ) + 2 = 9 so
√( a + 1/a ) = +√( 9 ) = +3
and therefore as
(a + 1/a)³ = (a³ + 1/a³) + 3(a + 1/a)
then
( 3 )³ = ( a³ + 1/a³ ) + 3( 3 ) or
27 = ( a³ + 1/a³ ) + 9 from which we have:
( a³ + 1/a³ ) = 27 - 9 = 18
Conclusion
( a³ + 1/a³ ) = 18