The value of left(frac{Delta^{2}}{E}right) x^ | vedclass

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(B) We know that $\Delta = E - 1$,so $\Delta^{2} = (E - 1)^{2} = E^{2} - 2E + 1$. Given expression is $\left(\frac{\Delta^{2}}{E}\right) x^{3} = \left

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$6 x$

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(B) We know that $\Delta = E - 1$,so $\Delta^{2} = (E - 1)^{2} = E^{2} - 2E + 1$. Given expression is $\left(\frac{\Delta^{2}}{E}\right) x^{3} = \left(\frac{E^{2} - 2E + 1}{E}\right) x^{3}$. This simplifies to $(E - 2 + E^{-1}) x^{3}$. Applying the operators: $E(x^{3}) = (x+1)^{3}$,$-2(x^{3}) = -2x^{3}$,and $E^{-1}(x^{3}) = (x-1)^{3}$. Summing these: $(x+1)^{3} - 2x^{3} + (x-1)^{3}$. Expanding the terms: $(x^{3} + 3x^{2} + 3x + 1) - 2x^{3} + (x^{3} - 3x^{2} + 3x - 1)$. Combining like terms: $(x^{3} - 2x^{3} + x^{3}) + (3x^{2} - 3x^{2}) + (3x + 3x) + (1 - 1) = 6x$.

The value of $\left(\frac{\Delta^{2}}{E}\right) x^{3}$ at $h=1$ is

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