If h = 1 then prove that (E-1 ∆) x3 = 3x2 – 3x + 1.

1.	If h = 1 then prove that (E-1 ∆) x3 = 3x2 – 3x + 1.
Answer» h = 1 To prove (E^-1 ∆) x³ = 3×2 – 3x + 1 L.H.S = (E^-1 ∆) x³ = E^-1 (∆x³ ) = E^-1 [(x + h)³ – x³ ] = E^-1 ( x + h)³ – E^-1 (x³) = (x – h + h)³ – (x – h)³ = x³ (x – h)³ But given h = 1 So(E^-1 ∆) x³ = x³ - (x – 1)³ = x³ – [x³ - 3x² + 3x – 1] = 3x² – 3x + 1 = RHS

Answer»

h = 1

To prove (E^-1 ∆) x³ = 3×2 – 3x + 1

L.H.S = (E^-1 ∆) x³ = E^-1 (∆x³ )

= E^-1 [(x + h)³ – x³ ]

= E^-1 ( x + h)³ – E^-1 (x³)

= (x – h + h)³ – (x – h)³

= x³ (x – h)³

But given h = 1

So(E^-1 ∆) x³ = x³ - (x – 1)³

= x³ – [x³ - 3x² + 3x – 1]

= 3x² – 3x + 1

= RHS

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