a2-3a+1=0 then find a3+1/a3
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72
Solution :
Given a² - 3a + 1 = 0
=> a² + 1 = 3a
divide each term by a , we get
=> a + 1/a = 3 -----( 1 )
***********************************
We know the algebraic identity :
x³ + y³ + 3xy( x + y ) = ( x + y )³
Or
x³ + y³ = ( x + y )³ - 3xy( x + y )
***********************************
Now ,
a³ + 1/a³ = ( a + 1/a )³ - 3×a×1/a( a + 1/a )
= 3³ - 3 × 3 [ from ( 1 ) ]
= 27 - 9
= 18
••••••
Given a² - 3a + 1 = 0
=> a² + 1 = 3a
divide each term by a , we get
=> a + 1/a = 3 -----( 1 )
***********************************
We know the algebraic identity :
x³ + y³ + 3xy( x + y ) = ( x + y )³
Or
x³ + y³ = ( x + y )³ - 3xy( x + y )
***********************************
Now ,
a³ + 1/a³ = ( a + 1/a )³ - 3×a×1/a( a + 1/a )
= 3³ - 3 × 3 [ from ( 1 ) ]
= 27 - 9
= 18
••••••
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40
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