Question

If m=(2-√3),then the value of (m6+m4+m2+1) /m3 is:

Answer 1

Heya,

given, (m⁶ + m⁴ + m² + 1)/m³

(m⁶/m³ + m⁴/m³ + m²/m³ + 1/m³)

(m³ + m + 1/m + 1/m³)

now,

m = (2 - √3)------------( 1 )

m³ = (8 - 3√3 - 6√3(2 - √3)

m³ = (8 - 3√3 - 12√3 + 18)

m³ = (26 - 15√3)-----------( 2 )

1/m = 1/(2 - √3) * (2 + √3)/(2 + √3)

1/m = (2 + √3)/(4 - 3)

1/m = (2 + √3)------------( 3 )

1/m³ = 1/(26 - 15√3) * (26 + 15√3)/(26 + 15√3)

1/m³ = (26 + 15√3)/(676 - 675)

1/m³ = (26 + 15√3)-------------( 4 )


adding------( 1 ) , --------( 2 ) , --------( 3 ) & --------( 4 )

(m + 1/m + m³ + 1/m³)

= (2 - √3 + 2 + √3 + 26 - 15√3 + 26 + 15√3)

= (4 + 52)

= 56

Hence, (m⁶ + m⁴ + m² + 1)/m³ = 56