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Question
If h = 1 then prove that (E–1Δ)x3 = 3x2 – 3x + 1
Solution
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
= R.H.S
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